2023-10-06 13:31:21 +08:00
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comments: true
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---
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# 14.1 初探动态规划
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2024-04-03 04:41:27 +08:00
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<u>动态规划(dynamic programming)</u>是一个重要的算法范式,它将一个问题分解为一系列更小的子问题,并通过存储子问题的解来避免重复计算,从而大幅提升时间效率。
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2023-10-06 13:31:21 +08:00
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在本节中,我们从一个经典例题入手,先给出它的暴力回溯解法,观察其中包含的重叠子问题,再逐步导出更高效的动态规划解法。
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!!! question "爬楼梯"
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2023-12-02 06:24:05 +08:00
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给定一个共有 $n$ 阶的楼梯,你每步可以上 $1$ 阶或者 $2$ 阶,请问有多少种方案可以爬到楼顶?
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2023-10-06 13:31:21 +08:00
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如图 14-1 所示,对于一个 $3$ 阶楼梯,共有 $3$ 种方案可以爬到楼顶。
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2023-11-09 05:13:48 +08:00
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![爬到第 3 阶的方案数量](intro_to_dynamic_programming.assets/climbing_stairs_example.png){ class="animation-figure" }
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2023-10-06 13:31:21 +08:00
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<p align="center"> 图 14-1 爬到第 3 阶的方案数量 </p>
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2023-12-02 06:24:05 +08:00
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本题的目标是求解方案数量,**我们可以考虑通过回溯来穷举所有可能性**。具体来说,将爬楼梯想象为一个多轮选择的过程:从地面出发,每轮选择上 $1$ 阶或 $2$ 阶,每当到达楼梯顶部时就将方案数量加 $1$ ,当越过楼梯顶部时就将其剪枝。代码如下所示:
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2023-10-06 13:31:21 +08:00
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=== "Python"
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```python title="climbing_stairs_backtrack.py"
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2023-10-06 14:10:18 +08:00
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def backtrack(choices: list[int], state: int, n: int, res: list[int]) -> int:
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"""回溯"""
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# 当爬到第 n 阶时,方案数量加 1
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if state == n:
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res[0] += 1
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# 遍历所有选择
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for choice in choices:
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# 剪枝:不允许越过第 n 阶
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if state + choice > n:
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2023-11-11 23:03:42 +08:00
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continue
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2023-10-06 14:10:18 +08:00
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# 尝试:做出选择,更新状态
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backtrack(choices, state + choice, n, res)
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# 回退
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def climbing_stairs_backtrack(n: int) -> int:
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"""爬楼梯:回溯"""
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2023-12-02 06:24:05 +08:00
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choices = [1, 2] # 可选择向上爬 1 阶或 2 阶
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2023-10-06 14:10:18 +08:00
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state = 0 # 从第 0 阶开始爬
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res = [0] # 使用 res[0] 记录方案数量
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backtrack(choices, state, n, res)
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return res[0]
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2023-10-06 13:31:21 +08:00
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```
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=== "C++"
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```cpp title="climbing_stairs_backtrack.cpp"
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2023-10-06 14:10:18 +08:00
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/* 回溯 */
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void backtrack(vector<int> &choices, int state, int n, vector<int> &res) {
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// 当爬到第 n 阶时,方案数量加 1
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if (state == n)
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res[0]++;
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// 遍历所有选择
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for (auto &choice : choices) {
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// 剪枝:不允许越过第 n 阶
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if (state + choice > n)
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2023-11-11 23:03:42 +08:00
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continue;
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2023-10-06 14:10:18 +08:00
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// 尝试:做出选择,更新状态
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backtrack(choices, state + choice, n, res);
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// 回退
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}
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}
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/* 爬楼梯:回溯 */
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int climbingStairsBacktrack(int n) {
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2023-12-02 06:24:05 +08:00
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vector<int> choices = {1, 2}; // 可选择向上爬 1 阶或 2 阶
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2023-10-06 14:10:18 +08:00
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int state = 0; // 从第 0 阶开始爬
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vector<int> res = {0}; // 使用 res[0] 记录方案数量
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backtrack(choices, state, n, res);
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return res[0];
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}
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2023-10-06 13:31:21 +08:00
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```
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=== "Java"
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```java title="climbing_stairs_backtrack.java"
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2023-10-06 14:10:18 +08:00
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/* 回溯 */
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void backtrack(List<Integer> choices, int state, int n, List<Integer> res) {
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// 当爬到第 n 阶时,方案数量加 1
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if (state == n)
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res.set(0, res.get(0) + 1);
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// 遍历所有选择
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for (Integer choice : choices) {
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// 剪枝:不允许越过第 n 阶
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if (state + choice > n)
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2023-11-11 23:03:42 +08:00
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continue;
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2023-10-06 14:10:18 +08:00
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// 尝试:做出选择,更新状态
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backtrack(choices, state + choice, n, res);
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// 回退
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}
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}
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/* 爬楼梯:回溯 */
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int climbingStairsBacktrack(int n) {
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2023-12-02 06:24:05 +08:00
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List<Integer> choices = Arrays.asList(1, 2); // 可选择向上爬 1 阶或 2 阶
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2023-10-06 14:10:18 +08:00
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int state = 0; // 从第 0 阶开始爬
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List<Integer> res = new ArrayList<>();
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res.add(0); // 使用 res[0] 记录方案数量
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backtrack(choices, state, n, res);
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return res.get(0);
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}
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2023-10-06 13:31:21 +08:00
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```
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=== "C#"
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```csharp title="climbing_stairs_backtrack.cs"
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2023-10-06 14:10:18 +08:00
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/* 回溯 */
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2023-10-08 01:43:28 +08:00
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void Backtrack(List<int> choices, int state, int n, List<int> res) {
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2023-10-06 14:10:18 +08:00
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// 当爬到第 n 阶时,方案数量加 1
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if (state == n)
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res[0]++;
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// 遍历所有选择
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foreach (int choice in choices) {
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// 剪枝:不允许越过第 n 阶
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if (state + choice > n)
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2023-11-11 23:03:42 +08:00
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continue;
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2023-10-06 14:10:18 +08:00
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// 尝试:做出选择,更新状态
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2023-10-08 01:43:28 +08:00
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Backtrack(choices, state + choice, n, res);
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2023-10-06 14:10:18 +08:00
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// 回退
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}
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}
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/* 爬楼梯:回溯 */
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2023-10-08 01:43:28 +08:00
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int ClimbingStairsBacktrack(int n) {
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2023-12-02 06:24:05 +08:00
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List<int> choices = [1, 2]; // 可选择向上爬 1 阶或 2 阶
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2023-10-06 14:10:18 +08:00
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int state = 0; // 从第 0 阶开始爬
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2023-11-27 02:32:06 +08:00
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List<int> res = [0]; // 使用 res[0] 记录方案数量
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2023-10-08 01:43:28 +08:00
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Backtrack(choices, state, n, res);
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2023-10-06 14:10:18 +08:00
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return res[0];
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}
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2023-10-06 13:31:21 +08:00
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```
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=== "Go"
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```go title="climbing_stairs_backtrack.go"
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2023-10-06 14:10:18 +08:00
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/* 回溯 */
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func backtrack(choices []int, state, n int, res []int) {
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// 当爬到第 n 阶时,方案数量加 1
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if state == n {
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res[0] = res[0] + 1
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}
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// 遍历所有选择
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for _, choice := range choices {
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// 剪枝:不允许越过第 n 阶
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if state+choice > n {
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2023-11-11 23:03:42 +08:00
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continue
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2023-10-06 14:10:18 +08:00
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}
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// 尝试:做出选择,更新状态
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backtrack(choices, state+choice, n, res)
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// 回退
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}
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}
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/* 爬楼梯:回溯 */
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func climbingStairsBacktrack(n int) int {
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2023-12-02 06:24:05 +08:00
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// 可选择向上爬 1 阶或 2 阶
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2023-10-06 14:10:18 +08:00
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choices := []int{1, 2}
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// 从第 0 阶开始爬
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state := 0
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res := make([]int, 1)
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// 使用 res[0] 记录方案数量
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res[0] = 0
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backtrack(choices, state, n, res)
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return res[0]
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}
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2023-10-06 13:31:21 +08:00
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```
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=== "Swift"
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```swift title="climbing_stairs_backtrack.swift"
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2023-10-06 14:10:18 +08:00
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/* 回溯 */
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func backtrack(choices: [Int], state: Int, n: Int, res: inout [Int]) {
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// 当爬到第 n 阶时,方案数量加 1
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if state == n {
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res[0] += 1
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}
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// 遍历所有选择
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for choice in choices {
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// 剪枝:不允许越过第 n 阶
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if state + choice > n {
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2023-11-11 23:03:42 +08:00
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continue
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2023-10-06 14:10:18 +08:00
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}
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2024-04-13 21:17:44 +08:00
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// 尝试:做出选择,更新状态
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2023-10-06 14:10:18 +08:00
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backtrack(choices: choices, state: state + choice, n: n, res: &res)
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2024-04-13 21:17:44 +08:00
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// 回退
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2023-10-06 14:10:18 +08:00
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}
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}
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/* 爬楼梯:回溯 */
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func climbingStairsBacktrack(n: Int) -> Int {
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2023-12-02 06:24:05 +08:00
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let choices = [1, 2] // 可选择向上爬 1 阶或 2 阶
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2023-10-06 14:10:18 +08:00
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let state = 0 // 从第 0 阶开始爬
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var res: [Int] = []
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res.append(0) // 使用 res[0] 记录方案数量
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backtrack(choices: choices, state: state, n: n, res: &res)
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return res[0]
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}
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2023-10-06 13:31:21 +08:00
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```
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=== "JS"
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```javascript title="climbing_stairs_backtrack.js"
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2023-10-06 14:10:18 +08:00
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/* 回溯 */
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function backtrack(choices, state, n, res) {
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// 当爬到第 n 阶时,方案数量加 1
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if (state === n) res.set(0, res.get(0) + 1);
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// 遍历所有选择
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for (const choice of choices) {
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// 剪枝:不允许越过第 n 阶
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2023-11-11 23:03:42 +08:00
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if (state + choice > n) continue;
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2023-10-06 14:10:18 +08:00
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// 尝试:做出选择,更新状态
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backtrack(choices, state + choice, n, res);
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// 回退
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}
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}
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/* 爬楼梯:回溯 */
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function climbingStairsBacktrack(n) {
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2023-12-02 06:24:05 +08:00
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const choices = [1, 2]; // 可选择向上爬 1 阶或 2 阶
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2023-10-06 14:10:18 +08:00
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const state = 0; // 从第 0 阶开始爬
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const res = new Map();
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res.set(0, 0); // 使用 res[0] 记录方案数量
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backtrack(choices, state, n, res);
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return res.get(0);
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}
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2023-10-06 13:31:21 +08:00
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```
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=== "TS"
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```typescript title="climbing_stairs_backtrack.ts"
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/* 回溯 */
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function backtrack(
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choices: number[],
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state: number,
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n: number,
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res: Map<0, any>
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): void {
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// 当爬到第 n 阶时,方案数量加 1
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if (state === n) res.set(0, res.get(0) + 1);
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// 遍历所有选择
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for (const choice of choices) {
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// 剪枝:不允许越过第 n 阶
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2023-11-11 23:03:42 +08:00
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if (state + choice > n) continue;
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2023-10-06 14:10:18 +08:00
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// 尝试:做出选择,更新状态
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backtrack(choices, state + choice, n, res);
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// 回退
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}
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}
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/* 爬楼梯:回溯 */
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function climbingStairsBacktrack(n: number): number {
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2023-12-02 06:24:05 +08:00
|
|
|
|
const choices = [1, 2]; // 可选择向上爬 1 阶或 2 阶
|
2023-10-06 14:10:18 +08:00
|
|
|
|
const state = 0; // 从第 0 阶开始爬
|
|
|
|
|
const res = new Map();
|
|
|
|
|
res.set(0, 0); // 使用 res[0] 记录方案数量
|
|
|
|
|
backtrack(choices, state, n, res);
|
|
|
|
|
return res.get(0);
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
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|
|
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|
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|
|
|
=== "Dart"
|
|
|
|
|
|
|
|
|
|
```dart title="climbing_stairs_backtrack.dart"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 回溯 */
|
|
|
|
|
void backtrack(List<int> choices, int state, int n, List<int> res) {
|
|
|
|
|
// 当爬到第 n 阶时,方案数量加 1
|
|
|
|
|
if (state == n) {
|
|
|
|
|
res[0]++;
|
|
|
|
|
}
|
|
|
|
|
// 遍历所有选择
|
|
|
|
|
for (int choice in choices) {
|
|
|
|
|
// 剪枝:不允许越过第 n 阶
|
2023-11-11 23:03:42 +08:00
|
|
|
|
if (state + choice > n) continue;
|
2023-10-06 14:10:18 +08:00
|
|
|
|
// 尝试:做出选择,更新状态
|
|
|
|
|
backtrack(choices, state + choice, n, res);
|
|
|
|
|
// 回退
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 爬楼梯:回溯 */
|
|
|
|
|
int climbingStairsBacktrack(int n) {
|
2023-12-02 06:24:05 +08:00
|
|
|
|
List<int> choices = [1, 2]; // 可选择向上爬 1 阶或 2 阶
|
2023-10-06 14:10:18 +08:00
|
|
|
|
int state = 0; // 从第 0 阶开始爬
|
|
|
|
|
List<int> res = [];
|
|
|
|
|
res.add(0); // 使用 res[0] 记录方案数量
|
|
|
|
|
backtrack(choices, state, n, res);
|
|
|
|
|
return res[0];
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Rust"
|
|
|
|
|
|
|
|
|
|
```rust title="climbing_stairs_backtrack.rs"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 回溯 */
|
|
|
|
|
fn backtrack(choices: &[i32], state: i32, n: i32, res: &mut [i32]) {
|
|
|
|
|
// 当爬到第 n 阶时,方案数量加 1
|
2024-03-18 03:11:07 +08:00
|
|
|
|
if state == n {
|
|
|
|
|
res[0] = res[0] + 1;
|
|
|
|
|
}
|
2023-10-06 14:10:18 +08:00
|
|
|
|
// 遍历所有选择
|
|
|
|
|
for &choice in choices {
|
|
|
|
|
// 剪枝:不允许越过第 n 阶
|
2024-03-18 03:11:07 +08:00
|
|
|
|
if state + choice > n {
|
|
|
|
|
continue;
|
|
|
|
|
}
|
2023-10-06 14:10:18 +08:00
|
|
|
|
// 尝试:做出选择,更新状态
|
|
|
|
|
backtrack(choices, state + choice, n, res);
|
|
|
|
|
// 回退
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 爬楼梯:回溯 */
|
|
|
|
|
fn climbing_stairs_backtrack(n: usize) -> i32 {
|
2024-03-18 03:11:07 +08:00
|
|
|
|
let choices = vec![1, 2]; // 可选择向上爬 1 阶或 2 阶
|
2023-10-06 14:10:18 +08:00
|
|
|
|
let state = 0; // 从第 0 阶开始爬
|
|
|
|
|
let mut res = Vec::new();
|
|
|
|
|
res.push(0); // 使用 res[0] 记录方案数量
|
|
|
|
|
backtrack(&choices, state, n as i32, &mut res);
|
|
|
|
|
res[0]
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
|
|
|
|
```c title="climbing_stairs_backtrack.c"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 回溯 */
|
|
|
|
|
void backtrack(int *choices, int state, int n, int *res, int len) {
|
|
|
|
|
// 当爬到第 n 阶时,方案数量加 1
|
|
|
|
|
if (state == n)
|
|
|
|
|
res[0]++;
|
|
|
|
|
// 遍历所有选择
|
|
|
|
|
for (int i = 0; i < len; i++) {
|
|
|
|
|
int choice = choices[i];
|
|
|
|
|
// 剪枝:不允许越过第 n 阶
|
|
|
|
|
if (state + choice > n)
|
2023-11-11 23:03:42 +08:00
|
|
|
|
continue;
|
2023-10-06 14:10:18 +08:00
|
|
|
|
// 尝试:做出选择,更新状态
|
|
|
|
|
backtrack(choices, state + choice, n, res, len);
|
|
|
|
|
// 回退
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 爬楼梯:回溯 */
|
|
|
|
|
int climbingStairsBacktrack(int n) {
|
2023-12-02 06:24:05 +08:00
|
|
|
|
int choices[2] = {1, 2}; // 可选择向上爬 1 阶或 2 阶
|
2023-10-06 14:10:18 +08:00
|
|
|
|
int state = 0; // 从第 0 阶开始爬
|
|
|
|
|
int *res = (int *)malloc(sizeof(int));
|
2023-10-18 02:16:55 +08:00
|
|
|
|
*res = 0; // 使用 res[0] 记录方案数量
|
2023-10-06 14:10:18 +08:00
|
|
|
|
int len = sizeof(choices) / sizeof(int);
|
|
|
|
|
backtrack(choices, state, n, res, len);
|
|
|
|
|
int result = *res;
|
|
|
|
|
free(res);
|
|
|
|
|
return result;
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
2024-03-25 22:43:12 +08:00
|
|
|
|
=== "Kotlin"
|
|
|
|
|
|
|
|
|
|
```kotlin title="climbing_stairs_backtrack.kt"
|
|
|
|
|
/* 回溯 */
|
|
|
|
|
fun backtrack(
|
2024-04-09 20:43:40 +08:00
|
|
|
|
choices: MutableList<Int>,
|
2024-03-25 22:43:12 +08:00
|
|
|
|
state: Int,
|
|
|
|
|
n: Int,
|
|
|
|
|
res: MutableList<Int>
|
|
|
|
|
) {
|
|
|
|
|
// 当爬到第 n 阶时,方案数量加 1
|
2024-04-09 20:43:40 +08:00
|
|
|
|
if (state == n)
|
|
|
|
|
res[0] = res[0] + 1
|
2024-03-25 22:43:12 +08:00
|
|
|
|
// 遍历所有选择
|
|
|
|
|
for (choice in choices) {
|
|
|
|
|
// 剪枝:不允许越过第 n 阶
|
|
|
|
|
if (state + choice > n) continue
|
|
|
|
|
// 尝试:做出选择,更新状态
|
|
|
|
|
backtrack(choices, state + choice, n, res)
|
|
|
|
|
// 回退
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 爬楼梯:回溯 */
|
|
|
|
|
fun climbingStairsBacktrack(n: Int): Int {
|
|
|
|
|
val choices = mutableListOf(1, 2) // 可选择向上爬 1 阶或 2 阶
|
|
|
|
|
val state = 0 // 从第 0 阶开始爬
|
2024-04-09 20:43:40 +08:00
|
|
|
|
val res = mutableListOf<Int>()
|
2024-03-25 22:43:12 +08:00
|
|
|
|
res.add(0) // 使用 res[0] 记录方案数量
|
|
|
|
|
backtrack(choices, state, n, res)
|
|
|
|
|
return res[0]
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
2024-03-31 03:53:04 +08:00
|
|
|
|
=== "Ruby"
|
|
|
|
|
|
|
|
|
|
```ruby title="climbing_stairs_backtrack.rb"
|
|
|
|
|
[class]{}-[func]{backtrack}
|
|
|
|
|
|
|
|
|
|
[class]{}-[func]{climbing_stairs_backtrack}
|
|
|
|
|
```
|
|
|
|
|
|
2023-10-06 13:31:21 +08:00
|
|
|
|
=== "Zig"
|
|
|
|
|
|
|
|
|
|
```zig title="climbing_stairs_backtrack.zig"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
// 回溯
|
|
|
|
|
fn backtrack(choices: []i32, state: i32, n: i32, res: std.ArrayList(i32)) void {
|
|
|
|
|
// 当爬到第 n 阶时,方案数量加 1
|
|
|
|
|
if (state == n) {
|
|
|
|
|
res.items[0] = res.items[0] + 1;
|
|
|
|
|
}
|
|
|
|
|
// 遍历所有选择
|
|
|
|
|
for (choices) |choice| {
|
|
|
|
|
// 剪枝:不允许越过第 n 阶
|
|
|
|
|
if (state + choice > n) {
|
2023-11-11 23:03:42 +08:00
|
|
|
|
continue;
|
2023-10-06 14:10:18 +08:00
|
|
|
|
}
|
|
|
|
|
// 尝试:做出选择,更新状态
|
|
|
|
|
backtrack(choices, state + choice, n, res);
|
|
|
|
|
// 回退
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
// 爬楼梯:回溯
|
|
|
|
|
fn climbingStairsBacktrack(n: usize) !i32 {
|
2023-12-02 06:24:05 +08:00
|
|
|
|
var choices = [_]i32{ 1, 2 }; // 可选择向上爬 1 阶或 2 阶
|
2023-10-06 14:10:18 +08:00
|
|
|
|
var state: i32 = 0; // 从第 0 阶开始爬
|
|
|
|
|
var res = std.ArrayList(i32).init(std.heap.page_allocator);
|
|
|
|
|
defer res.deinit();
|
|
|
|
|
try res.append(0); // 使用 res[0] 记录方案数量
|
|
|
|
|
backtrack(&choices, state, @intCast(n), res);
|
|
|
|
|
return res.items[0];
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
2024-01-07 23:42:54 +08:00
|
|
|
|
??? pythontutor "可视化运行"
|
|
|
|
|
|
2024-01-09 16:00:24 +08:00
|
|
|
|
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20backtrack%28choices%3A%20list%5Bint%5D,%20state%3A%20int,%20n%3A%20int,%20res%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%9B%9E%E6%BA%AF%22%22%22%0A%20%20%20%20%23%20%E5%BD%93%E7%88%AC%E5%88%B0%E7%AC%AC%20n%20%E9%98%B6%E6%97%B6%EF%BC%8C%E6%96%B9%E6%A1%88%E6%95%B0%E9%87%8F%E5%8A%A0%201%0A%20%20%20%20if%20state%20%3D%3D%20n%3A%0A%20%20%20%20%20%20%20%20res%5B0%5D%20%2B%3D%201%0A%20%20%20%20%23%20%E9%81%8D%E5%8E%86%E6%89%80%E6%9C%89%E9%80%89%E6%8B%A9%0A%20%20%20%20for%20choice%20in%20choices%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%EF%BC%9A%E4%B8%8D%E5%85%81%E8%AE%B8%E8%B6%8A%E8%BF%87%E7%AC%AC%20n%20%E9%98%B6%0A%20%20%20%20%20%20%20%20if%20state%20%2B%20choice%20%3E%20n%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20continue%0A%20%20%20%20%20%20%20%20%23%20%E5%B0%9D%E8%AF%95%EF%BC%9A%E5%81%9A%E5%87%BA%E9%80%89%E6%8B%A9%EF%BC%8C%E6%9B%B4%E6%96%B0%E7%8A%B6%E6%80%81%0A%20%20%20%20%20%20%20%20backtrack%28choices,%20state%20%2B%20choice,%20n,%20res%29%0A%20%20%20%20%20%20%20%20%23%20%E5%9B%9E%E9%80%80%0A%0A%0Adef%20climbing_stairs_backtrack%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E5%9B%9E%E6%BA%AF%22%22%22%0A%20%20%20%20choices%20%3D%20%5B1,%202%5D%20%20%23%20%E5%8F%AF%E9%80%89%E6%8B%A9%E5%90%91%E4%B8%8A%E7%88%AC%201%20%E9%98%B6%E6%88%96%202%20%E9%98%B6%0A%20%20%20%20state%20%3D%200%20%20%23%20%E4%BB%8E%E7%AC%AC%200%20%E9%98%B6%E5%BC%80%E5%A7%8B%E7%88%AC%0A%20%20%20%20res%20%3D%20%5B0%5D%20%20%23%20%E4%BD%BF%E7%94%A8%20res%5B0%5D%20%E8%AE%B0%E5%BD%95%E6%96%B9%E6%A1%88%E6%95%B0%E9%87%8F%0A%20%20%20%20backtrack%28choices,%20state,%20n,%20res%29%0A%20%20%20%20return%20res%5B0%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%204%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_backtrack%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
|
|
|
|
|
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20backtrack%28choices%3A%20list%5Bint%5D,%20state%3A%20int,%20n%3A%20int,%20res%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%9B%9E%E6%BA%AF%22%22%22%0A%20%20%20%20%23%20%E5%BD%93%E7%88%AC%E5%88%B0%E7%AC%AC%20n%20%E9%98%B6%E6%97%B6%EF%BC%8C%E6%96%B9%E6%A1%88%E6%95%B0%E9%87%8F%E5%8A%A0%201%0A%20%20%20%20if%20state%20%3D%3D%20n%3A%0A%20%20%20%20%20%20%20%20res%5B0%5D%20%2B%3D%201%0A%20%20%20%20%23%20%E9%81%8D%E5%8E%86%E6%89%80%E6%9C%89%E9%80%89%E6%8B%A9%0A%20%20%20%20for%20choice%20in%20choices%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%EF%BC%9A%E4%B8%8D%E5%85%81%E8%AE%B8%E8%B6%8A%E8%BF%87%E7%AC%AC%20n%20%E9%98%B6%0A%20%20%20%20%20%20%20%20if%20state%20%2B%20choice%20%3E%20n%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20continue%0A%20%20%20%20%20%20%20%20%23%20%E5%B0%9D%E8%AF%95%EF%BC%9A%E5%81%9A%E5%87%BA%E9%80%89%E6%8B%A9%EF%BC%8C%E6%9B%B4%E6%96%B0%E7%8A%B6%E6%80%81%0A%20%20%20%20%20%20%20%20backtrack%28choices,%20state%20%2B%20choice,%20n,%20res%29%0A%20%20%20%20%20%20%20%20%23%20%E5%9B%9E%E9%80%80%0A%0A%0Adef%20climbing_stairs_backtrack%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E5%9B%9E%E6%BA%AF%22%22%22%0A%20%20%20%20choices%20%3D%20%5B1,%202%5D%20%20%23%20%E5%8F%AF%E9%80%89%E6%8B%A9%E5%90%91%E4%B8%8A%E7%88%AC%201%20%E9%98%B6%E6%88%96%202%20%E9%98%B6%0A%20%20%20%20state%20%3D%200%20%20%23%20%E4%BB%8E%E7%AC%AC%200%20%E9%98%B6%E5%BC%80%E5%A7%8B%E7%88%AC%0A%20%20%20%20res%20%3D%20%5B0%5D%20%20%23%20%E4%BD%BF%E7%94%A8%20res%5B0%5D%20%E8%AE%B0%E5%BD%95%E6%96%B9%E6%A1%88%E6%95%B0%E9%87%8F%0A%20%20%20%20backtrack%28choices,%20state,%20n,%20res%29%0A%20%20%20%20return%20res%5B0%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%204%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_backtrack%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全屏观看 ></a></div>
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2024-01-07 23:42:54 +08:00
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2023-10-06 13:31:21 +08:00
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## 14.1.1 方法一:暴力搜索
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2023-12-02 06:24:05 +08:00
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回溯算法通常并不显式地对问题进行拆解,而是将求解问题看作一系列决策步骤,通过试探和剪枝,搜索所有可能的解。
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2023-10-06 13:31:21 +08:00
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2023-12-02 06:24:05 +08:00
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我们可以尝试从问题分解的角度分析这道题。设爬到第 $i$ 阶共有 $dp[i]$ 种方案,那么 $dp[i]$ 就是原问题,其子问题包括:
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2023-10-06 13:31:21 +08:00
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$$
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dp[i-1], dp[i-2], \dots, dp[2], dp[1]
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$$
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2023-12-02 06:24:05 +08:00
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由于每轮只能上 $1$ 阶或 $2$ 阶,因此当我们站在第 $i$ 阶楼梯上时,上一轮只可能站在第 $i - 1$ 阶或第 $i - 2$ 阶上。换句话说,我们只能从第 $i -1$ 阶或第 $i - 2$ 阶迈向第 $i$ 阶。
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2023-10-06 13:31:21 +08:00
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由此便可得出一个重要推论:**爬到第 $i - 1$ 阶的方案数加上爬到第 $i - 2$ 阶的方案数就等于爬到第 $i$ 阶的方案数**。公式如下:
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$$
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dp[i] = dp[i-1] + dp[i-2]
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$$
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这意味着在爬楼梯问题中,各个子问题之间存在递推关系,**原问题的解可以由子问题的解构建得来**。图 14-2 展示了该递推关系。
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2023-11-09 05:13:48 +08:00
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![方案数量递推关系](intro_to_dynamic_programming.assets/climbing_stairs_state_transfer.png){ class="animation-figure" }
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2023-10-06 13:31:21 +08:00
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<p align="center"> 图 14-2 方案数量递推关系 </p>
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我们可以根据递推公式得到暴力搜索解法。以 $dp[n]$ 为起始点,**递归地将一个较大问题拆解为两个较小问题的和**,直至到达最小子问题 $dp[1]$ 和 $dp[2]$ 时返回。其中,最小子问题的解是已知的,即 $dp[1] = 1$、$dp[2] = 2$ ,表示爬到第 $1$、$2$ 阶分别有 $1$、$2$ 种方案。
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2023-12-02 06:24:05 +08:00
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观察以下代码,它和标准回溯代码都属于深度优先搜索,但更加简洁:
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2023-10-06 13:31:21 +08:00
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=== "Python"
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```python title="climbing_stairs_dfs.py"
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2023-10-06 14:10:18 +08:00
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def dfs(i: int) -> int:
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"""搜索"""
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# 已知 dp[1] 和 dp[2] ,返回之
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if i == 1 or i == 2:
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return i
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# dp[i] = dp[i-1] + dp[i-2]
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count = dfs(i - 1) + dfs(i - 2)
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return count
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def climbing_stairs_dfs(n: int) -> int:
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"""爬楼梯:搜索"""
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return dfs(n)
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2023-10-06 13:31:21 +08:00
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```
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=== "C++"
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```cpp title="climbing_stairs_dfs.cpp"
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2023-10-06 14:10:18 +08:00
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/* 搜索 */
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int dfs(int i) {
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// 已知 dp[1] 和 dp[2] ,返回之
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if (i == 1 || i == 2)
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return i;
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// dp[i] = dp[i-1] + dp[i-2]
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int count = dfs(i - 1) + dfs(i - 2);
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return count;
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}
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/* 爬楼梯:搜索 */
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int climbingStairsDFS(int n) {
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return dfs(n);
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}
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2023-10-06 13:31:21 +08:00
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```
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=== "Java"
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```java title="climbing_stairs_dfs.java"
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2023-10-06 14:10:18 +08:00
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/* 搜索 */
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int dfs(int i) {
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// 已知 dp[1] 和 dp[2] ,返回之
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if (i == 1 || i == 2)
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return i;
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// dp[i] = dp[i-1] + dp[i-2]
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int count = dfs(i - 1) + dfs(i - 2);
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return count;
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}
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/* 爬楼梯:搜索 */
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int climbingStairsDFS(int n) {
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return dfs(n);
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}
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2023-10-06 13:31:21 +08:00
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```
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=== "C#"
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```csharp title="climbing_stairs_dfs.cs"
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2023-10-06 14:10:18 +08:00
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/* 搜索 */
|
2023-10-15 21:18:21 +08:00
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int DFS(int i) {
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2023-10-06 14:10:18 +08:00
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// 已知 dp[1] 和 dp[2] ,返回之
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if (i == 1 || i == 2)
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return i;
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// dp[i] = dp[i-1] + dp[i-2]
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2023-10-15 21:18:21 +08:00
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int count = DFS(i - 1) + DFS(i - 2);
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2023-10-06 14:10:18 +08:00
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return count;
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}
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/* 爬楼梯:搜索 */
|
2023-10-08 01:43:28 +08:00
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int ClimbingStairsDFS(int n) {
|
2023-10-15 21:18:21 +08:00
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return DFS(n);
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2023-10-06 14:10:18 +08:00
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}
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2023-10-06 13:31:21 +08:00
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```
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|
=== "Go"
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|
```go title="climbing_stairs_dfs.go"
|
2023-10-06 14:10:18 +08:00
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/* 搜索 */
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func dfs(i int) int {
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|
// 已知 dp[1] 和 dp[2] ,返回之
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if i == 1 || i == 2 {
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return i
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}
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// dp[i] = dp[i-1] + dp[i-2]
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count := dfs(i-1) + dfs(i-2)
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return count
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}
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/* 爬楼梯:搜索 */
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func climbingStairsDFS(n int) int {
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return dfs(n)
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}
|
2023-10-06 13:31:21 +08:00
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```
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|
=== "Swift"
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|
```swift title="climbing_stairs_dfs.swift"
|
2023-10-06 14:10:18 +08:00
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|
/* 搜索 */
|
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|
func dfs(i: Int) -> Int {
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|
// 已知 dp[1] 和 dp[2] ,返回之
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if i == 1 || i == 2 {
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return i
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}
|
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|
// dp[i] = dp[i-1] + dp[i-2]
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let count = dfs(i: i - 1) + dfs(i: i - 2)
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return count
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}
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|
|
/* 爬楼梯:搜索 */
|
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|
|
func climbingStairsDFS(n: Int) -> Int {
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dfs(i: n)
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}
|
2023-10-06 13:31:21 +08:00
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|
```
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|
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|
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|
=== "JS"
|
|
|
|
|
|
|
|
|
|
```javascript title="climbing_stairs_dfs.js"
|
2023-10-06 14:10:18 +08:00
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|
|
/* 搜索 */
|
|
|
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|
function dfs(i) {
|
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|
// 已知 dp[1] 和 dp[2] ,返回之
|
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|
if (i === 1 || i === 2) return i;
|
|
|
|
|
// dp[i] = dp[i-1] + dp[i-2]
|
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|
|
const count = dfs(i - 1) + dfs(i - 2);
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|
return count;
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|
}
|
|
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|
|
|
|
|
|
|
/* 爬楼梯:搜索 */
|
|
|
|
|
function climbingStairsDFS(n) {
|
|
|
|
|
return dfs(n);
|
|
|
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|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "TS"
|
|
|
|
|
|
|
|
|
|
```typescript title="climbing_stairs_dfs.ts"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 搜索 */
|
|
|
|
|
function dfs(i: number): number {
|
|
|
|
|
// 已知 dp[1] 和 dp[2] ,返回之
|
|
|
|
|
if (i === 1 || i === 2) return i;
|
|
|
|
|
// dp[i] = dp[i-1] + dp[i-2]
|
|
|
|
|
const count = dfs(i - 1) + dfs(i - 2);
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 爬楼梯:搜索 */
|
|
|
|
|
function climbingStairsDFS(n: number): number {
|
|
|
|
|
return dfs(n);
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
|
|
|
|
```dart title="climbing_stairs_dfs.dart"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 搜索 */
|
|
|
|
|
int dfs(int i) {
|
|
|
|
|
// 已知 dp[1] 和 dp[2] ,返回之
|
|
|
|
|
if (i == 1 || i == 2) return i;
|
|
|
|
|
// dp[i] = dp[i-1] + dp[i-2]
|
|
|
|
|
int count = dfs(i - 1) + dfs(i - 2);
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 爬楼梯:搜索 */
|
|
|
|
|
int climbingStairsDFS(int n) {
|
|
|
|
|
return dfs(n);
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Rust"
|
|
|
|
|
|
|
|
|
|
```rust title="climbing_stairs_dfs.rs"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 搜索 */
|
|
|
|
|
fn dfs(i: usize) -> i32 {
|
|
|
|
|
// 已知 dp[1] 和 dp[2] ,返回之
|
2024-03-18 03:11:07 +08:00
|
|
|
|
if i == 1 || i == 2 {
|
|
|
|
|
return i as i32;
|
|
|
|
|
}
|
2023-10-06 14:10:18 +08:00
|
|
|
|
// dp[i] = dp[i-1] + dp[i-2]
|
|
|
|
|
let count = dfs(i - 1) + dfs(i - 2);
|
|
|
|
|
count
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 爬楼梯:搜索 */
|
|
|
|
|
fn climbing_stairs_dfs(n: usize) -> i32 {
|
2024-03-18 03:11:07 +08:00
|
|
|
|
dfs(n)
|
2023-10-06 14:10:18 +08:00
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
|
|
|
|
```c title="climbing_stairs_dfs.c"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 搜索 */
|
|
|
|
|
int dfs(int i) {
|
|
|
|
|
// 已知 dp[1] 和 dp[2] ,返回之
|
|
|
|
|
if (i == 1 || i == 2)
|
|
|
|
|
return i;
|
|
|
|
|
// dp[i] = dp[i-1] + dp[i-2]
|
|
|
|
|
int count = dfs(i - 1) + dfs(i - 2);
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 爬楼梯:搜索 */
|
|
|
|
|
int climbingStairsDFS(int n) {
|
|
|
|
|
return dfs(n);
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
2024-03-25 22:43:12 +08:00
|
|
|
|
=== "Kotlin"
|
|
|
|
|
|
|
|
|
|
```kotlin title="climbing_stairs_dfs.kt"
|
|
|
|
|
/* 搜索 */
|
|
|
|
|
fun dfs(i: Int): Int {
|
|
|
|
|
// 已知 dp[1] 和 dp[2] ,返回之
|
|
|
|
|
if (i == 1 || i == 2) return i
|
|
|
|
|
// dp[i] = dp[i-1] + dp[i-2]
|
|
|
|
|
val count = dfs(i - 1) + dfs(i - 2)
|
|
|
|
|
return count
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 爬楼梯:搜索 */
|
|
|
|
|
fun climbingStairsDFS(n: Int): Int {
|
|
|
|
|
return dfs(n)
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
2024-03-31 03:53:04 +08:00
|
|
|
|
=== "Ruby"
|
|
|
|
|
|
|
|
|
|
```ruby title="climbing_stairs_dfs.rb"
|
|
|
|
|
[class]{}-[func]{dfs}
|
|
|
|
|
|
|
|
|
|
[class]{}-[func]{climbing_stairs_dfs}
|
|
|
|
|
```
|
|
|
|
|
|
2023-10-06 13:31:21 +08:00
|
|
|
|
=== "Zig"
|
|
|
|
|
|
|
|
|
|
```zig title="climbing_stairs_dfs.zig"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
// 搜索
|
|
|
|
|
fn dfs(i: usize) i32 {
|
|
|
|
|
// 已知 dp[1] 和 dp[2] ,返回之
|
|
|
|
|
if (i == 1 or i == 2) {
|
|
|
|
|
return @intCast(i);
|
|
|
|
|
}
|
|
|
|
|
// dp[i] = dp[i-1] + dp[i-2]
|
|
|
|
|
var count = dfs(i - 1) + dfs(i - 2);
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
// 爬楼梯:搜索
|
|
|
|
|
fn climbingStairsDFS(comptime n: usize) i32 {
|
|
|
|
|
return dfs(n);
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
2024-01-07 23:42:54 +08:00
|
|
|
|
??? pythontutor "可视化运行"
|
|
|
|
|
|
2024-01-09 16:00:24 +08:00
|
|
|
|
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20dfs%28i%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20%E5%B7%B2%E7%9F%A5%20dp%5B1%5D%20%E5%92%8C%20dp%5B2%5D%20%EF%BC%8C%E8%BF%94%E5%9B%9E%E4%B9%8B%0A%20%20%20%20if%20i%20%3D%3D%201%20or%20i%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20i%0A%20%20%20%20%23%20dp%5Bi%5D%20%3D%20dp%5Bi-1%5D%20%2B%20dp%5Bi-2%5D%0A%20%20%20%20count%20%3D%20dfs%28i%20-%201%29%20%2B%20dfs%28i%20-%202%29%0A%20%20%20%20return%20count%0A%0A%0Adef%20climbing_stairs_dfs%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20return%20dfs%28n%29%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_dfs%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
|
|
|
|
|
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20dfs%28i%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20%E5%B7%B2%E7%9F%A5%20dp%5B1%5D%20%E5%92%8C%20dp%5B2%5D%20%EF%BC%8C%E8%BF%94%E5%9B%9E%E4%B9%8B%0A%20%20%20%20if%20i%20%3D%3D%201%20or%20i%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20i%0A%20%20%20%20%23%20dp%5Bi%5D%20%3D%20dp%5Bi-1%5D%20%2B%20dp%5Bi-2%5D%0A%20%20%20%20count%20%3D%20dfs%28i%20-%201%29%20%2B%20dfs%28i%20-%202%29%0A%20%20%20%20return%20count%0A%0A%0Adef%20climbing_stairs_dfs%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20return%20dfs%28n%29%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_dfs%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全屏观看 ></a></div>
|
2024-01-07 23:42:54 +08:00
|
|
|
|
|
2023-10-06 13:31:21 +08:00
|
|
|
|
图 14-3 展示了暴力搜索形成的递归树。对于问题 $dp[n]$ ,其递归树的深度为 $n$ ,时间复杂度为 $O(2^n)$ 。指数阶属于爆炸式增长,如果我们输入一个比较大的 $n$ ,则会陷入漫长的等待之中。
|
|
|
|
|
|
2023-11-09 05:13:48 +08:00
|
|
|
|
![爬楼梯对应递归树](intro_to_dynamic_programming.assets/climbing_stairs_dfs_tree.png){ class="animation-figure" }
|
2023-10-06 13:31:21 +08:00
|
|
|
|
|
|
|
|
|
<p align="center"> 图 14-3 爬楼梯对应递归树 </p>
|
|
|
|
|
|
2023-12-02 06:24:05 +08:00
|
|
|
|
观察图 14-3 ,**指数阶的时间复杂度是“重叠子问题”导致的**。例如 $dp[9]$ 被分解为 $dp[8]$ 和 $dp[7]$ ,$dp[8]$ 被分解为 $dp[7]$ 和 $dp[6]$ ,两者都包含子问题 $dp[7]$ 。
|
2023-10-06 13:31:21 +08:00
|
|
|
|
|
2023-12-28 17:18:37 +08:00
|
|
|
|
以此类推,子问题中包含更小的重叠子问题,子子孙孙无穷尽也。绝大部分计算资源都浪费在这些重叠的子问题上。
|
2023-10-06 13:31:21 +08:00
|
|
|
|
|
|
|
|
|
## 14.1.2 方法二:记忆化搜索
|
|
|
|
|
|
|
|
|
|
为了提升算法效率,**我们希望所有的重叠子问题都只被计算一次**。为此,我们声明一个数组 `mem` 来记录每个子问题的解,并在搜索过程中将重叠子问题剪枝。
|
|
|
|
|
|
|
|
|
|
1. 当首次计算 $dp[i]$ 时,我们将其记录至 `mem[i]` ,以便之后使用。
|
|
|
|
|
2. 当再次需要计算 $dp[i]$ 时,我们便可直接从 `mem[i]` 中获取结果,从而避免重复计算该子问题。
|
|
|
|
|
|
2023-12-02 06:24:05 +08:00
|
|
|
|
代码如下所示:
|
|
|
|
|
|
2023-10-06 13:31:21 +08:00
|
|
|
|
=== "Python"
|
|
|
|
|
|
|
|
|
|
```python title="climbing_stairs_dfs_mem.py"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
def dfs(i: int, mem: list[int]) -> int:
|
|
|
|
|
"""记忆化搜索"""
|
|
|
|
|
# 已知 dp[1] 和 dp[2] ,返回之
|
|
|
|
|
if i == 1 or i == 2:
|
|
|
|
|
return i
|
|
|
|
|
# 若存在记录 dp[i] ,则直接返回之
|
|
|
|
|
if mem[i] != -1:
|
|
|
|
|
return mem[i]
|
|
|
|
|
# dp[i] = dp[i-1] + dp[i-2]
|
|
|
|
|
count = dfs(i - 1, mem) + dfs(i - 2, mem)
|
|
|
|
|
# 记录 dp[i]
|
|
|
|
|
mem[i] = count
|
|
|
|
|
return count
|
|
|
|
|
|
|
|
|
|
def climbing_stairs_dfs_mem(n: int) -> int:
|
|
|
|
|
"""爬楼梯:记忆化搜索"""
|
|
|
|
|
# mem[i] 记录爬到第 i 阶的方案总数,-1 代表无记录
|
|
|
|
|
mem = [-1] * (n + 1)
|
|
|
|
|
return dfs(n, mem)
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
|
|
|
|
```cpp title="climbing_stairs_dfs_mem.cpp"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 记忆化搜索 */
|
|
|
|
|
int dfs(int i, vector<int> &mem) {
|
|
|
|
|
// 已知 dp[1] 和 dp[2] ,返回之
|
|
|
|
|
if (i == 1 || i == 2)
|
|
|
|
|
return i;
|
|
|
|
|
// 若存在记录 dp[i] ,则直接返回之
|
|
|
|
|
if (mem[i] != -1)
|
|
|
|
|
return mem[i];
|
|
|
|
|
// dp[i] = dp[i-1] + dp[i-2]
|
|
|
|
|
int count = dfs(i - 1, mem) + dfs(i - 2, mem);
|
|
|
|
|
// 记录 dp[i]
|
|
|
|
|
mem[i] = count;
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 爬楼梯:记忆化搜索 */
|
|
|
|
|
int climbingStairsDFSMem(int n) {
|
|
|
|
|
// mem[i] 记录爬到第 i 阶的方案总数,-1 代表无记录
|
|
|
|
|
vector<int> mem(n + 1, -1);
|
|
|
|
|
return dfs(n, mem);
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
|
|
|
|
```java title="climbing_stairs_dfs_mem.java"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 记忆化搜索 */
|
|
|
|
|
int dfs(int i, int[] mem) {
|
|
|
|
|
// 已知 dp[1] 和 dp[2] ,返回之
|
|
|
|
|
if (i == 1 || i == 2)
|
|
|
|
|
return i;
|
|
|
|
|
// 若存在记录 dp[i] ,则直接返回之
|
|
|
|
|
if (mem[i] != -1)
|
|
|
|
|
return mem[i];
|
|
|
|
|
// dp[i] = dp[i-1] + dp[i-2]
|
|
|
|
|
int count = dfs(i - 1, mem) + dfs(i - 2, mem);
|
|
|
|
|
// 记录 dp[i]
|
|
|
|
|
mem[i] = count;
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 爬楼梯:记忆化搜索 */
|
|
|
|
|
int climbingStairsDFSMem(int n) {
|
|
|
|
|
// mem[i] 记录爬到第 i 阶的方案总数,-1 代表无记录
|
|
|
|
|
int[] mem = new int[n + 1];
|
|
|
|
|
Arrays.fill(mem, -1);
|
|
|
|
|
return dfs(n, mem);
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
|
|
|
|
```csharp title="climbing_stairs_dfs_mem.cs"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 记忆化搜索 */
|
2023-10-15 21:18:21 +08:00
|
|
|
|
int DFS(int i, int[] mem) {
|
2023-10-06 14:10:18 +08:00
|
|
|
|
// 已知 dp[1] 和 dp[2] ,返回之
|
|
|
|
|
if (i == 1 || i == 2)
|
|
|
|
|
return i;
|
|
|
|
|
// 若存在记录 dp[i] ,则直接返回之
|
|
|
|
|
if (mem[i] != -1)
|
|
|
|
|
return mem[i];
|
|
|
|
|
// dp[i] = dp[i-1] + dp[i-2]
|
2023-10-15 21:18:21 +08:00
|
|
|
|
int count = DFS(i - 1, mem) + DFS(i - 2, mem);
|
2023-10-06 14:10:18 +08:00
|
|
|
|
// 记录 dp[i]
|
|
|
|
|
mem[i] = count;
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 爬楼梯:记忆化搜索 */
|
2023-10-08 01:43:28 +08:00
|
|
|
|
int ClimbingStairsDFSMem(int n) {
|
2023-10-06 14:10:18 +08:00
|
|
|
|
// mem[i] 记录爬到第 i 阶的方案总数,-1 代表无记录
|
|
|
|
|
int[] mem = new int[n + 1];
|
|
|
|
|
Array.Fill(mem, -1);
|
2023-10-15 21:18:21 +08:00
|
|
|
|
return DFS(n, mem);
|
2023-10-06 14:10:18 +08:00
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
|
|
|
|
```go title="climbing_stairs_dfs_mem.go"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 记忆化搜索 */
|
|
|
|
|
func dfsMem(i int, mem []int) int {
|
|
|
|
|
// 已知 dp[1] 和 dp[2] ,返回之
|
|
|
|
|
if i == 1 || i == 2 {
|
|
|
|
|
return i
|
|
|
|
|
}
|
|
|
|
|
// 若存在记录 dp[i] ,则直接返回之
|
|
|
|
|
if mem[i] != -1 {
|
|
|
|
|
return mem[i]
|
|
|
|
|
}
|
|
|
|
|
// dp[i] = dp[i-1] + dp[i-2]
|
|
|
|
|
count := dfsMem(i-1, mem) + dfsMem(i-2, mem)
|
|
|
|
|
// 记录 dp[i]
|
|
|
|
|
mem[i] = count
|
|
|
|
|
return count
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 爬楼梯:记忆化搜索 */
|
|
|
|
|
func climbingStairsDFSMem(n int) int {
|
|
|
|
|
// mem[i] 记录爬到第 i 阶的方案总数,-1 代表无记录
|
|
|
|
|
mem := make([]int, n+1)
|
|
|
|
|
for i := range mem {
|
|
|
|
|
mem[i] = -1
|
|
|
|
|
}
|
|
|
|
|
return dfsMem(n, mem)
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
|
|
|
|
```swift title="climbing_stairs_dfs_mem.swift"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 记忆化搜索 */
|
|
|
|
|
func dfs(i: Int, mem: inout [Int]) -> Int {
|
|
|
|
|
// 已知 dp[1] 和 dp[2] ,返回之
|
|
|
|
|
if i == 1 || i == 2 {
|
|
|
|
|
return i
|
|
|
|
|
}
|
|
|
|
|
// 若存在记录 dp[i] ,则直接返回之
|
|
|
|
|
if mem[i] != -1 {
|
|
|
|
|
return mem[i]
|
|
|
|
|
}
|
|
|
|
|
// dp[i] = dp[i-1] + dp[i-2]
|
|
|
|
|
let count = dfs(i: i - 1, mem: &mem) + dfs(i: i - 2, mem: &mem)
|
|
|
|
|
// 记录 dp[i]
|
|
|
|
|
mem[i] = count
|
|
|
|
|
return count
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 爬楼梯:记忆化搜索 */
|
|
|
|
|
func climbingStairsDFSMem(n: Int) -> Int {
|
|
|
|
|
// mem[i] 记录爬到第 i 阶的方案总数,-1 代表无记录
|
|
|
|
|
var mem = Array(repeating: -1, count: n + 1)
|
|
|
|
|
return dfs(i: n, mem: &mem)
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "JS"
|
|
|
|
|
|
|
|
|
|
```javascript title="climbing_stairs_dfs_mem.js"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 记忆化搜索 */
|
|
|
|
|
function dfs(i, mem) {
|
|
|
|
|
// 已知 dp[1] 和 dp[2] ,返回之
|
|
|
|
|
if (i === 1 || i === 2) return i;
|
|
|
|
|
// 若存在记录 dp[i] ,则直接返回之
|
|
|
|
|
if (mem[i] != -1) return mem[i];
|
|
|
|
|
// dp[i] = dp[i-1] + dp[i-2]
|
|
|
|
|
const count = dfs(i - 1, mem) + dfs(i - 2, mem);
|
|
|
|
|
// 记录 dp[i]
|
|
|
|
|
mem[i] = count;
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 爬楼梯:记忆化搜索 */
|
|
|
|
|
function climbingStairsDFSMem(n) {
|
|
|
|
|
// mem[i] 记录爬到第 i 阶的方案总数,-1 代表无记录
|
|
|
|
|
const mem = new Array(n + 1).fill(-1);
|
|
|
|
|
return dfs(n, mem);
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "TS"
|
|
|
|
|
|
|
|
|
|
```typescript title="climbing_stairs_dfs_mem.ts"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 记忆化搜索 */
|
|
|
|
|
function dfs(i: number, mem: number[]): number {
|
|
|
|
|
// 已知 dp[1] 和 dp[2] ,返回之
|
|
|
|
|
if (i === 1 || i === 2) return i;
|
|
|
|
|
// 若存在记录 dp[i] ,则直接返回之
|
|
|
|
|
if (mem[i] != -1) return mem[i];
|
|
|
|
|
// dp[i] = dp[i-1] + dp[i-2]
|
|
|
|
|
const count = dfs(i - 1, mem) + dfs(i - 2, mem);
|
|
|
|
|
// 记录 dp[i]
|
|
|
|
|
mem[i] = count;
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 爬楼梯:记忆化搜索 */
|
|
|
|
|
function climbingStairsDFSMem(n: number): number {
|
|
|
|
|
// mem[i] 记录爬到第 i 阶的方案总数,-1 代表无记录
|
|
|
|
|
const mem = new Array(n + 1).fill(-1);
|
|
|
|
|
return dfs(n, mem);
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
|
|
|
|
```dart title="climbing_stairs_dfs_mem.dart"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 记忆化搜索 */
|
|
|
|
|
int dfs(int i, List<int> mem) {
|
|
|
|
|
// 已知 dp[1] 和 dp[2] ,返回之
|
|
|
|
|
if (i == 1 || i == 2) return i;
|
|
|
|
|
// 若存在记录 dp[i] ,则直接返回之
|
|
|
|
|
if (mem[i] != -1) return mem[i];
|
|
|
|
|
// dp[i] = dp[i-1] + dp[i-2]
|
|
|
|
|
int count = dfs(i - 1, mem) + dfs(i - 2, mem);
|
|
|
|
|
// 记录 dp[i]
|
|
|
|
|
mem[i] = count;
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 爬楼梯:记忆化搜索 */
|
|
|
|
|
int climbingStairsDFSMem(int n) {
|
|
|
|
|
// mem[i] 记录爬到第 i 阶的方案总数,-1 代表无记录
|
|
|
|
|
List<int> mem = List.filled(n + 1, -1);
|
|
|
|
|
return dfs(n, mem);
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Rust"
|
|
|
|
|
|
|
|
|
|
```rust title="climbing_stairs_dfs_mem.rs"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 记忆化搜索 */
|
|
|
|
|
fn dfs(i: usize, mem: &mut [i32]) -> i32 {
|
|
|
|
|
// 已知 dp[1] 和 dp[2] ,返回之
|
2024-03-18 03:11:07 +08:00
|
|
|
|
if i == 1 || i == 2 {
|
|
|
|
|
return i as i32;
|
|
|
|
|
}
|
2023-10-06 14:10:18 +08:00
|
|
|
|
// 若存在记录 dp[i] ,则直接返回之
|
2024-03-18 03:11:07 +08:00
|
|
|
|
if mem[i] != -1 {
|
|
|
|
|
return mem[i];
|
|
|
|
|
}
|
2023-10-06 14:10:18 +08:00
|
|
|
|
// dp[i] = dp[i-1] + dp[i-2]
|
|
|
|
|
let count = dfs(i - 1, mem) + dfs(i - 2, mem);
|
|
|
|
|
// 记录 dp[i]
|
|
|
|
|
mem[i] = count;
|
|
|
|
|
count
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 爬楼梯:记忆化搜索 */
|
|
|
|
|
fn climbing_stairs_dfs_mem(n: usize) -> i32 {
|
|
|
|
|
// mem[i] 记录爬到第 i 阶的方案总数,-1 代表无记录
|
|
|
|
|
let mut mem = vec![-1; n + 1];
|
|
|
|
|
dfs(n, &mut mem)
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
|
|
|
|
```c title="climbing_stairs_dfs_mem.c"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 记忆化搜索 */
|
|
|
|
|
int dfs(int i, int *mem) {
|
|
|
|
|
// 已知 dp[1] 和 dp[2] ,返回之
|
|
|
|
|
if (i == 1 || i == 2)
|
|
|
|
|
return i;
|
|
|
|
|
// 若存在记录 dp[i] ,则直接返回之
|
|
|
|
|
if (mem[i] != -1)
|
|
|
|
|
return mem[i];
|
|
|
|
|
// dp[i] = dp[i-1] + dp[i-2]
|
|
|
|
|
int count = dfs(i - 1, mem) + dfs(i - 2, mem);
|
|
|
|
|
// 记录 dp[i]
|
|
|
|
|
mem[i] = count;
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 爬楼梯:记忆化搜索 */
|
|
|
|
|
int climbingStairsDFSMem(int n) {
|
|
|
|
|
// mem[i] 记录爬到第 i 阶的方案总数,-1 代表无记录
|
|
|
|
|
int *mem = (int *)malloc((n + 1) * sizeof(int));
|
|
|
|
|
for (int i = 0; i <= n; i++) {
|
|
|
|
|
mem[i] = -1;
|
|
|
|
|
}
|
|
|
|
|
int result = dfs(n, mem);
|
|
|
|
|
free(mem);
|
|
|
|
|
return result;
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
2024-03-25 22:43:12 +08:00
|
|
|
|
=== "Kotlin"
|
|
|
|
|
|
|
|
|
|
```kotlin title="climbing_stairs_dfs_mem.kt"
|
|
|
|
|
/* 记忆化搜索 */
|
|
|
|
|
fun dfs(i: Int, mem: IntArray): Int {
|
|
|
|
|
// 已知 dp[1] 和 dp[2] ,返回之
|
|
|
|
|
if (i == 1 || i == 2) return i
|
|
|
|
|
// 若存在记录 dp[i] ,则直接返回之
|
|
|
|
|
if (mem[i] != -1) return mem[i]
|
|
|
|
|
// dp[i] = dp[i-1] + dp[i-2]
|
|
|
|
|
val count = dfs(i - 1, mem) + dfs(i - 2, mem)
|
|
|
|
|
// 记录 dp[i]
|
|
|
|
|
mem[i] = count
|
|
|
|
|
return count
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
/* 爬楼梯:记忆化搜索 */
|
|
|
|
|
fun climbingStairsDFSMem(n: Int): Int {
|
|
|
|
|
// mem[i] 记录爬到第 i 阶的方案总数,-1 代表无记录
|
|
|
|
|
val mem = IntArray(n + 1)
|
2024-04-09 20:43:40 +08:00
|
|
|
|
mem.fill(-1)
|
2024-03-25 22:43:12 +08:00
|
|
|
|
return dfs(n, mem)
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
2024-03-31 03:53:04 +08:00
|
|
|
|
=== "Ruby"
|
|
|
|
|
|
|
|
|
|
```ruby title="climbing_stairs_dfs_mem.rb"
|
|
|
|
|
[class]{}-[func]{dfs}
|
|
|
|
|
|
|
|
|
|
[class]{}-[func]{climbing_stairs_dfs_mem}
|
|
|
|
|
```
|
|
|
|
|
|
2023-10-06 13:31:21 +08:00
|
|
|
|
=== "Zig"
|
|
|
|
|
|
|
|
|
|
```zig title="climbing_stairs_dfs_mem.zig"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
// 记忆化搜索
|
|
|
|
|
fn dfs(i: usize, mem: []i32) i32 {
|
|
|
|
|
// 已知 dp[1] 和 dp[2] ,返回之
|
|
|
|
|
if (i == 1 or i == 2) {
|
|
|
|
|
return @intCast(i);
|
|
|
|
|
}
|
|
|
|
|
// 若存在记录 dp[i] ,则直接返回之
|
|
|
|
|
if (mem[i] != -1) {
|
|
|
|
|
return mem[i];
|
|
|
|
|
}
|
|
|
|
|
// dp[i] = dp[i-1] + dp[i-2]
|
|
|
|
|
var count = dfs(i - 1, mem) + dfs(i - 2, mem);
|
|
|
|
|
// 记录 dp[i]
|
|
|
|
|
mem[i] = count;
|
|
|
|
|
return count;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
// 爬楼梯:记忆化搜索
|
|
|
|
|
fn climbingStairsDFSMem(comptime n: usize) i32 {
|
|
|
|
|
// mem[i] 记录爬到第 i 阶的方案总数,-1 代表无记录
|
|
|
|
|
var mem = [_]i32{ -1 } ** (n + 1);
|
|
|
|
|
return dfs(n, &mem);
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
2024-01-07 23:42:54 +08:00
|
|
|
|
??? pythontutor "可视化运行"
|
|
|
|
|
|
2024-01-09 16:00:24 +08:00
|
|
|
|
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20dfs%28i%3A%20int,%20mem%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E8%AE%B0%E5%BF%86%E5%8C%96%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20%E5%B7%B2%E7%9F%A5%20dp%5B1%5D%20%E5%92%8C%20dp%5B2%5D%20%EF%BC%8C%E8%BF%94%E5%9B%9E%E4%B9%8B%0A%20%20%20%20if%20i%20%3D%3D%201%20or%20i%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20i%0A%20%20%20%20%23%20%E8%8B%A5%E5%AD%98%E5%9C%A8%E8%AE%B0%E5%BD%95%20dp%5Bi%5D%20%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E8%BF%94%E5%9B%9E%E4%B9%8B%0A%20%20%20%20if%20mem%5Bi%5D%20!%3D%20-1%3A%0A%20%20%20%20%20%20%20%20return%20mem%5Bi%5D%0A%20%20%20%20%23%20dp%5Bi%5D%20%3D%20dp%5Bi-1%5D%20%2B%20dp%5Bi-2%5D%0A%20%20%20%20count%20%3D%20dfs%28i%20-%201,%20mem%29%20%2B%20dfs%28i%20-%202,%20mem%29%0A%20%20%20%20%23%20%E8%AE%B0%E5%BD%95%20dp%5Bi%5D%0A%20%20%20%20mem%5Bi%5D%20%3D%20count%0A%20%20%20%20return%20count%0A%0A%0Adef%20climbing_stairs_dfs_mem%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E8%AE%B0%E5%BF%86%E5%8C%96%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20mem%5Bi%5D%20%E8%AE%B0%E5%BD%95%E7%88%AC%E5%88%B0%E7%AC%AC%20i%20%E9%98%B6%E7%9A%84%E6%96%B9%E6%A1%88%E6%80%BB%E6%95%B0%EF%BC%8C-1%20%E4%BB%A3%E8%A1%A8%E6%97%A0%E8%AE%B0%E5%BD%95%0A%20%20%20%20mem%20%3D%20%5B-1%5D%20*%20%28n%20%2B%201%29%0A%20%20%20%20return%20dfs%28n,%20mem%29%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_dfs_mem%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
|
|
|
|
|
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20dfs%28i%3A%20int,%20mem%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E8%AE%B0%E5%BF%86%E5%8C%96%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20%E5%B7%B2%E7%9F%A5%20dp%5B1%5D%20%E5%92%8C%20dp%5B2%5D%20%EF%BC%8C%E8%BF%94%E5%9B%9E%E4%B9%8B%0A%20%20%20%20if%20i%20%3D%3D%201%20or%20i%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20i%0A%20%20%20%20%23%20%E8%8B%A5%E5%AD%98%E5%9C%A8%E8%AE%B0%E5%BD%95%20dp%5Bi%5D%20%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E8%BF%94%E5%9B%9E%E4%B9%8B%0A%20%20%20%20if%20mem%5Bi%5D%20!%3D%20-1%3A%0A%20%20%20%20%20%20%20%20return%20mem%5Bi%5D%0A%20%20%20%20%23%20dp%5Bi%5D%20%3D%20dp%5Bi-1%5D%20%2B%20dp%5Bi-2%5D%0A%20%20%20%20count%20%3D%20dfs%28i%20-%201,%20mem%29%20%2B%20dfs%28i%20-%202,%20mem%29%0A%20%20%20%20%23%20%E8%AE%B0%E5%BD%95%20dp%5Bi%5D%0A%20%20%20%20mem%5Bi%5D%20%3D%20count%0A%20%20%20%20return%20count%0A%0A%0Adef%20climbing_stairs_dfs_mem%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E8%AE%B0%E5%BF%86%E5%8C%96%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20mem%5Bi%5D%20%E8%AE%B0%E5%BD%95%E7%88%AC%E5%88%B0%E7%AC%AC%20i%20%E9%98%B6%E7%9A%84%E6%96%B9%E6%A1%88%E6%80%BB%E6%95%B0%EF%BC%8C-1%20%E4%BB%A3%E8%A1%A8%E6%97%A0%E8%AE%B0%E5%BD%95%0A%20%20%20%20mem%20%3D%20%5B-1%5D%20*%20%28n%20%2B%201%29%0A%20%20%20%20return%20dfs%28n,%20mem%29%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_dfs_mem%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全屏观看 ></a></div>
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2024-01-07 23:42:54 +08:00
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2023-12-02 06:24:05 +08:00
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观察图 14-4 ,**经过记忆化处理后,所有重叠子问题都只需计算一次,时间复杂度优化至 $O(n)$** ,这是一个巨大的飞跃。
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2023-10-06 13:31:21 +08:00
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2023-11-09 05:13:48 +08:00
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![记忆化搜索对应递归树](intro_to_dynamic_programming.assets/climbing_stairs_dfs_memo_tree.png){ class="animation-figure" }
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2023-10-06 13:31:21 +08:00
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<p align="center"> 图 14-4 记忆化搜索对应递归树 </p>
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## 14.1.3 方法三:动态规划
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2023-12-02 06:24:05 +08:00
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**记忆化搜索是一种“从顶至底”的方法**:我们从原问题(根节点)开始,递归地将较大子问题分解为较小子问题,直至解已知的最小子问题(叶节点)。之后,通过回溯逐层收集子问题的解,构建出原问题的解。
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2023-10-06 13:31:21 +08:00
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与之相反,**动态规划是一种“从底至顶”的方法**:从最小子问题的解开始,迭代地构建更大子问题的解,直至得到原问题的解。
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2023-12-02 06:24:05 +08:00
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由于动态规划不包含回溯过程,因此只需使用循环迭代实现,无须使用递归。在以下代码中,我们初始化一个数组 `dp` 来存储子问题的解,它起到了与记忆化搜索中数组 `mem` 相同的记录作用:
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2023-10-06 13:31:21 +08:00
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=== "Python"
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```python title="climbing_stairs_dp.py"
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2023-10-06 14:10:18 +08:00
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def climbing_stairs_dp(n: int) -> int:
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"""爬楼梯:动态规划"""
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if n == 1 or n == 2:
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return n
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# 初始化 dp 表,用于存储子问题的解
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dp = [0] * (n + 1)
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# 初始状态:预设最小子问题的解
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dp[1], dp[2] = 1, 2
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# 状态转移:从较小子问题逐步求解较大子问题
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for i in range(3, n + 1):
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dp[i] = dp[i - 1] + dp[i - 2]
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return dp[n]
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2023-10-06 13:31:21 +08:00
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```
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=== "C++"
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```cpp title="climbing_stairs_dp.cpp"
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2023-10-06 14:10:18 +08:00
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/* 爬楼梯:动态规划 */
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int climbingStairsDP(int n) {
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if (n == 1 || n == 2)
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return n;
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// 初始化 dp 表,用于存储子问题的解
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vector<int> dp(n + 1);
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// 初始状态:预设最小子问题的解
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dp[1] = 1;
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dp[2] = 2;
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// 状态转移:从较小子问题逐步求解较大子问题
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for (int i = 3; i <= n; i++) {
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dp[i] = dp[i - 1] + dp[i - 2];
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}
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return dp[n];
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}
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2023-10-06 13:31:21 +08:00
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```
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=== "Java"
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```java title="climbing_stairs_dp.java"
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2023-10-06 14:10:18 +08:00
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/* 爬楼梯:动态规划 */
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int climbingStairsDP(int n) {
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if (n == 1 || n == 2)
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return n;
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// 初始化 dp 表,用于存储子问题的解
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int[] dp = new int[n + 1];
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// 初始状态:预设最小子问题的解
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dp[1] = 1;
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dp[2] = 2;
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// 状态转移:从较小子问题逐步求解较大子问题
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for (int i = 3; i <= n; i++) {
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dp[i] = dp[i - 1] + dp[i - 2];
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}
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return dp[n];
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}
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2023-10-06 13:31:21 +08:00
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```
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=== "C#"
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```csharp title="climbing_stairs_dp.cs"
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2023-10-06 14:10:18 +08:00
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/* 爬楼梯:动态规划 */
|
2023-10-08 01:43:28 +08:00
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int ClimbingStairsDP(int n) {
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2023-10-06 14:10:18 +08:00
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if (n == 1 || n == 2)
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return n;
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|
// 初始化 dp 表,用于存储子问题的解
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int[] dp = new int[n + 1];
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// 初始状态:预设最小子问题的解
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dp[1] = 1;
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dp[2] = 2;
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// 状态转移:从较小子问题逐步求解较大子问题
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for (int i = 3; i <= n; i++) {
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dp[i] = dp[i - 1] + dp[i - 2];
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}
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return dp[n];
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}
|
2023-10-06 13:31:21 +08:00
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```
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=== "Go"
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|
```go title="climbing_stairs_dp.go"
|
2023-10-06 14:10:18 +08:00
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|
/* 爬楼梯:动态规划 */
|
|
|
|
|
func climbingStairsDP(n int) int {
|
|
|
|
|
if n == 1 || n == 2 {
|
|
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|
return n
|
|
|
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|
}
|
|
|
|
|
// 初始化 dp 表,用于存储子问题的解
|
|
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|
dp := make([]int, n+1)
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|
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|
// 初始状态:预设最小子问题的解
|
|
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|
dp[1] = 1
|
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|
dp[2] = 2
|
|
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|
// 状态转移:从较小子问题逐步求解较大子问题
|
|
|
|
|
for i := 3; i <= n; i++ {
|
|
|
|
|
dp[i] = dp[i-1] + dp[i-2]
|
|
|
|
|
}
|
|
|
|
|
return dp[n]
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
|
|
|
|
```swift title="climbing_stairs_dp.swift"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 爬楼梯:动态规划 */
|
|
|
|
|
func climbingStairsDP(n: Int) -> Int {
|
|
|
|
|
if n == 1 || n == 2 {
|
|
|
|
|
return n
|
|
|
|
|
}
|
|
|
|
|
// 初始化 dp 表,用于存储子问题的解
|
|
|
|
|
var dp = Array(repeating: 0, count: n + 1)
|
|
|
|
|
// 初始状态:预设最小子问题的解
|
|
|
|
|
dp[1] = 1
|
|
|
|
|
dp[2] = 2
|
|
|
|
|
// 状态转移:从较小子问题逐步求解较大子问题
|
2024-03-21 04:22:07 +08:00
|
|
|
|
for i in 3 ... n {
|
2023-10-06 14:10:18 +08:00
|
|
|
|
dp[i] = dp[i - 1] + dp[i - 2]
|
|
|
|
|
}
|
|
|
|
|
return dp[n]
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "JS"
|
|
|
|
|
|
|
|
|
|
```javascript title="climbing_stairs_dp.js"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 爬楼梯:动态规划 */
|
|
|
|
|
function climbingStairsDP(n) {
|
|
|
|
|
if (n === 1 || n === 2) return n;
|
|
|
|
|
// 初始化 dp 表,用于存储子问题的解
|
|
|
|
|
const dp = new Array(n + 1).fill(-1);
|
|
|
|
|
// 初始状态:预设最小子问题的解
|
|
|
|
|
dp[1] = 1;
|
|
|
|
|
dp[2] = 2;
|
|
|
|
|
// 状态转移:从较小子问题逐步求解较大子问题
|
|
|
|
|
for (let i = 3; i <= n; i++) {
|
|
|
|
|
dp[i] = dp[i - 1] + dp[i - 2];
|
|
|
|
|
}
|
|
|
|
|
return dp[n];
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "TS"
|
|
|
|
|
|
|
|
|
|
```typescript title="climbing_stairs_dp.ts"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 爬楼梯:动态规划 */
|
|
|
|
|
function climbingStairsDP(n: number): number {
|
|
|
|
|
if (n === 1 || n === 2) return n;
|
|
|
|
|
// 初始化 dp 表,用于存储子问题的解
|
|
|
|
|
const dp = new Array(n + 1).fill(-1);
|
|
|
|
|
// 初始状态:预设最小子问题的解
|
|
|
|
|
dp[1] = 1;
|
|
|
|
|
dp[2] = 2;
|
|
|
|
|
// 状态转移:从较小子问题逐步求解较大子问题
|
|
|
|
|
for (let i = 3; i <= n; i++) {
|
|
|
|
|
dp[i] = dp[i - 1] + dp[i - 2];
|
|
|
|
|
}
|
|
|
|
|
return dp[n];
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
|
|
|
|
```dart title="climbing_stairs_dp.dart"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 爬楼梯:动态规划 */
|
|
|
|
|
int climbingStairsDP(int n) {
|
|
|
|
|
if (n == 1 || n == 2) return n;
|
|
|
|
|
// 初始化 dp 表,用于存储子问题的解
|
|
|
|
|
List<int> dp = List.filled(n + 1, 0);
|
|
|
|
|
// 初始状态:预设最小子问题的解
|
|
|
|
|
dp[1] = 1;
|
|
|
|
|
dp[2] = 2;
|
|
|
|
|
// 状态转移:从较小子问题逐步求解较大子问题
|
|
|
|
|
for (int i = 3; i <= n; i++) {
|
|
|
|
|
dp[i] = dp[i - 1] + dp[i - 2];
|
|
|
|
|
}
|
|
|
|
|
return dp[n];
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Rust"
|
|
|
|
|
|
|
|
|
|
```rust title="climbing_stairs_dp.rs"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 爬楼梯:动态规划 */
|
|
|
|
|
fn climbing_stairs_dp(n: usize) -> i32 {
|
|
|
|
|
// 已知 dp[1] 和 dp[2] ,返回之
|
2024-03-18 03:11:07 +08:00
|
|
|
|
if n == 1 || n == 2 {
|
|
|
|
|
return n as i32;
|
|
|
|
|
}
|
2023-10-06 14:10:18 +08:00
|
|
|
|
// 初始化 dp 表,用于存储子问题的解
|
|
|
|
|
let mut dp = vec![-1; n + 1];
|
|
|
|
|
// 初始状态:预设最小子问题的解
|
|
|
|
|
dp[1] = 1;
|
|
|
|
|
dp[2] = 2;
|
|
|
|
|
// 状态转移:从较小子问题逐步求解较大子问题
|
|
|
|
|
for i in 3..=n {
|
|
|
|
|
dp[i] = dp[i - 1] + dp[i - 2];
|
|
|
|
|
}
|
|
|
|
|
dp[n]
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
|
|
|
|
```c title="climbing_stairs_dp.c"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 爬楼梯:动态规划 */
|
|
|
|
|
int climbingStairsDP(int n) {
|
|
|
|
|
if (n == 1 || n == 2)
|
|
|
|
|
return n;
|
|
|
|
|
// 初始化 dp 表,用于存储子问题的解
|
|
|
|
|
int *dp = (int *)malloc((n + 1) * sizeof(int));
|
|
|
|
|
// 初始状态:预设最小子问题的解
|
|
|
|
|
dp[1] = 1;
|
|
|
|
|
dp[2] = 2;
|
|
|
|
|
// 状态转移:从较小子问题逐步求解较大子问题
|
|
|
|
|
for (int i = 3; i <= n; i++) {
|
|
|
|
|
dp[i] = dp[i - 1] + dp[i - 2];
|
|
|
|
|
}
|
|
|
|
|
int result = dp[n];
|
|
|
|
|
free(dp);
|
|
|
|
|
return result;
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
2024-03-25 22:43:12 +08:00
|
|
|
|
=== "Kotlin"
|
|
|
|
|
|
|
|
|
|
```kotlin title="climbing_stairs_dp.kt"
|
|
|
|
|
/* 爬楼梯:动态规划 */
|
|
|
|
|
fun climbingStairsDP(n: Int): Int {
|
|
|
|
|
if (n == 1 || n == 2) return n
|
|
|
|
|
// 初始化 dp 表,用于存储子问题的解
|
|
|
|
|
val dp = IntArray(n + 1)
|
|
|
|
|
// 初始状态:预设最小子问题的解
|
|
|
|
|
dp[1] = 1
|
|
|
|
|
dp[2] = 2
|
|
|
|
|
// 状态转移:从较小子问题逐步求解较大子问题
|
|
|
|
|
for (i in 3..n) {
|
|
|
|
|
dp[i] = dp[i - 1] + dp[i - 2]
|
|
|
|
|
}
|
|
|
|
|
return dp[n]
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
2024-03-31 03:53:04 +08:00
|
|
|
|
=== "Ruby"
|
|
|
|
|
|
|
|
|
|
```ruby title="climbing_stairs_dp.rb"
|
|
|
|
|
[class]{}-[func]{climbing_stairs_dp}
|
|
|
|
|
```
|
|
|
|
|
|
2023-10-06 13:31:21 +08:00
|
|
|
|
=== "Zig"
|
|
|
|
|
|
|
|
|
|
```zig title="climbing_stairs_dp.zig"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
// 爬楼梯:动态规划
|
|
|
|
|
fn climbingStairsDP(comptime n: usize) i32 {
|
|
|
|
|
// 已知 dp[1] 和 dp[2] ,返回之
|
|
|
|
|
if (n == 1 or n == 2) {
|
|
|
|
|
return @intCast(n);
|
|
|
|
|
}
|
|
|
|
|
// 初始化 dp 表,用于存储子问题的解
|
|
|
|
|
var dp = [_]i32{-1} ** (n + 1);
|
|
|
|
|
// 初始状态:预设最小子问题的解
|
|
|
|
|
dp[1] = 1;
|
|
|
|
|
dp[2] = 2;
|
|
|
|
|
// 状态转移:从较小子问题逐步求解较大子问题
|
|
|
|
|
for (3..n + 1) |i| {
|
|
|
|
|
dp[i] = dp[i - 1] + dp[i - 2];
|
|
|
|
|
}
|
|
|
|
|
return dp[n];
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
2024-01-07 23:42:54 +08:00
|
|
|
|
??? pythontutor "可视化运行"
|
|
|
|
|
|
2024-01-09 16:00:24 +08:00
|
|
|
|
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20climbing_stairs_dp%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20n%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%EF%BC%8C%E7%94%A8%E4%BA%8E%E5%AD%98%E5%82%A8%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%20%3D%20%5B0%5D%20*%20%28n%20%2B%201%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E7%8A%B6%E6%80%81%EF%BC%9A%E9%A2%84%E8%AE%BE%E6%9C%80%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%5B1%5D,%20dp%5B2%5D%20%3D%201,%202%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E4%BB%8E%E8%BE%83%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E9%80%90%E6%AD%A5%E6%B1%82%E8%A7%A3%E8%BE%83%E5%A4%A7%E5%AD%90%E9%97%AE%E9%A2%98%0A%20%20%20%20for%20i%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%20%3D%20dp%5Bi%20-%201%5D%20%2B%20dp%5Bi%20-%202%5D%0A%20%20%20%20return%20dp%5Bn%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_dp%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
|
|
|
|
|
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20climbing_stairs_dp%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20n%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%EF%BC%8C%E7%94%A8%E4%BA%8E%E5%AD%98%E5%82%A8%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%20%3D%20%5B0%5D%20*%20%28n%20%2B%201%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E7%8A%B6%E6%80%81%EF%BC%9A%E9%A2%84%E8%AE%BE%E6%9C%80%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%5B1%5D,%20dp%5B2%5D%20%3D%201,%202%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E4%BB%8E%E8%BE%83%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E9%80%90%E6%AD%A5%E6%B1%82%E8%A7%A3%E8%BE%83%E5%A4%A7%E5%AD%90%E9%97%AE%E9%A2%98%0A%20%20%20%20for%20i%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%20%3D%20dp%5Bi%20-%201%5D%20%2B%20dp%5Bi%20-%202%5D%0A%20%20%20%20return%20dp%5Bn%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_dp%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全屏观看 ></a></div>
|
2024-01-07 23:42:54 +08:00
|
|
|
|
|
2023-10-06 13:31:21 +08:00
|
|
|
|
图 14-5 模拟了以上代码的执行过程。
|
|
|
|
|
|
2023-11-09 05:13:48 +08:00
|
|
|
|
![爬楼梯的动态规划过程](intro_to_dynamic_programming.assets/climbing_stairs_dp.png){ class="animation-figure" }
|
2023-10-06 13:31:21 +08:00
|
|
|
|
|
|
|
|
|
<p align="center"> 图 14-5 爬楼梯的动态规划过程 </p>
|
|
|
|
|
|
2023-12-02 06:24:05 +08:00
|
|
|
|
与回溯算法一样,动态规划也使用“状态”概念来表示问题求解的特定阶段,每个状态都对应一个子问题以及相应的局部最优解。例如,爬楼梯问题的状态定义为当前所在楼梯阶数 $i$ 。
|
2023-10-06 13:31:21 +08:00
|
|
|
|
|
|
|
|
|
根据以上内容,我们可以总结出动态规划的常用术语。
|
|
|
|
|
|
2024-04-03 15:44:08 +08:00
|
|
|
|
- 将数组 `dp` 称为 <u>dp 表</u>,$dp[i]$ 表示状态 $i$ 对应子问题的解。
|
2024-04-03 04:41:27 +08:00
|
|
|
|
- 将最小子问题对应的状态(第 $1$ 阶和第 $2$ 阶楼梯)称为<u>初始状态</u>。
|
|
|
|
|
- 将递推公式 $dp[i] = dp[i-1] + dp[i-2]$ 称为<u>状态转移方程</u>。
|
2023-10-06 13:31:21 +08:00
|
|
|
|
|
|
|
|
|
## 14.1.4 空间优化
|
|
|
|
|
|
2023-12-02 06:24:05 +08:00
|
|
|
|
细心的读者可能发现了,**由于 $dp[i]$ 只与 $dp[i-1]$ 和 $dp[i-2]$ 有关,因此我们无须使用一个数组 `dp` 来存储所有子问题的解**,而只需两个变量滚动前进即可。代码如下所示:
|
2023-10-06 13:31:21 +08:00
|
|
|
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
|
|
|
|
```python title="climbing_stairs_dp.py"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
def climbing_stairs_dp_comp(n: int) -> int:
|
|
|
|
|
"""爬楼梯:空间优化后的动态规划"""
|
|
|
|
|
if n == 1 or n == 2:
|
|
|
|
|
return n
|
|
|
|
|
a, b = 1, 2
|
|
|
|
|
for _ in range(3, n + 1):
|
|
|
|
|
a, b = b, a + b
|
|
|
|
|
return b
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
|
|
|
|
```cpp title="climbing_stairs_dp.cpp"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 爬楼梯:空间优化后的动态规划 */
|
|
|
|
|
int climbingStairsDPComp(int n) {
|
|
|
|
|
if (n == 1 || n == 2)
|
|
|
|
|
return n;
|
|
|
|
|
int a = 1, b = 2;
|
|
|
|
|
for (int i = 3; i <= n; i++) {
|
|
|
|
|
int tmp = b;
|
|
|
|
|
b = a + b;
|
|
|
|
|
a = tmp;
|
|
|
|
|
}
|
|
|
|
|
return b;
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
|
|
|
|
```java title="climbing_stairs_dp.java"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 爬楼梯:空间优化后的动态规划 */
|
|
|
|
|
int climbingStairsDPComp(int n) {
|
|
|
|
|
if (n == 1 || n == 2)
|
|
|
|
|
return n;
|
|
|
|
|
int a = 1, b = 2;
|
|
|
|
|
for (int i = 3; i <= n; i++) {
|
|
|
|
|
int tmp = b;
|
|
|
|
|
b = a + b;
|
|
|
|
|
a = tmp;
|
|
|
|
|
}
|
|
|
|
|
return b;
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
|
|
|
|
```csharp title="climbing_stairs_dp.cs"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 爬楼梯:空间优化后的动态规划 */
|
2023-10-08 01:43:28 +08:00
|
|
|
|
int ClimbingStairsDPComp(int n) {
|
2023-10-06 14:10:18 +08:00
|
|
|
|
if (n == 1 || n == 2)
|
|
|
|
|
return n;
|
|
|
|
|
int a = 1, b = 2;
|
|
|
|
|
for (int i = 3; i <= n; i++) {
|
|
|
|
|
int tmp = b;
|
|
|
|
|
b = a + b;
|
|
|
|
|
a = tmp;
|
|
|
|
|
}
|
|
|
|
|
return b;
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
|
|
|
|
```go title="climbing_stairs_dp.go"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 爬楼梯:空间优化后的动态规划 */
|
|
|
|
|
func climbingStairsDPComp(n int) int {
|
|
|
|
|
if n == 1 || n == 2 {
|
|
|
|
|
return n
|
|
|
|
|
}
|
|
|
|
|
a, b := 1, 2
|
|
|
|
|
// 状态转移:从较小子问题逐步求解较大子问题
|
|
|
|
|
for i := 3; i <= n; i++ {
|
|
|
|
|
a, b = b, a+b
|
|
|
|
|
}
|
|
|
|
|
return b
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
|
|
|
|
```swift title="climbing_stairs_dp.swift"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 爬楼梯:空间优化后的动态规划 */
|
|
|
|
|
func climbingStairsDPComp(n: Int) -> Int {
|
|
|
|
|
if n == 1 || n == 2 {
|
|
|
|
|
return n
|
|
|
|
|
}
|
|
|
|
|
var a = 1
|
|
|
|
|
var b = 2
|
2024-03-21 04:22:07 +08:00
|
|
|
|
for _ in 3 ... n {
|
2023-10-06 14:10:18 +08:00
|
|
|
|
(a, b) = (b, a + b)
|
|
|
|
|
}
|
|
|
|
|
return b
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "JS"
|
|
|
|
|
|
|
|
|
|
```javascript title="climbing_stairs_dp.js"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 爬楼梯:空间优化后的动态规划 */
|
|
|
|
|
function climbingStairsDPComp(n) {
|
|
|
|
|
if (n === 1 || n === 2) return n;
|
|
|
|
|
let a = 1,
|
|
|
|
|
b = 2;
|
|
|
|
|
for (let i = 3; i <= n; i++) {
|
|
|
|
|
const tmp = b;
|
|
|
|
|
b = a + b;
|
|
|
|
|
a = tmp;
|
|
|
|
|
}
|
|
|
|
|
return b;
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "TS"
|
|
|
|
|
|
|
|
|
|
```typescript title="climbing_stairs_dp.ts"
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2023-10-06 14:10:18 +08:00
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/* 爬楼梯:空间优化后的动态规划 */
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function climbingStairsDPComp(n: number): number {
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if (n === 1 || n === 2) return n;
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let a = 1,
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b = 2;
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for (let i = 3; i <= n; i++) {
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const tmp = b;
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b = a + b;
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a = tmp;
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}
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return b;
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}
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2023-10-06 13:31:21 +08:00
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```
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=== "Dart"
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```dart title="climbing_stairs_dp.dart"
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2023-10-06 14:10:18 +08:00
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/* 爬楼梯:空间优化后的动态规划 */
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int climbingStairsDPComp(int n) {
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if (n == 1 || n == 2) return n;
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int a = 1, b = 2;
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for (int i = 3; i <= n; i++) {
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int tmp = b;
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b = a + b;
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a = tmp;
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}
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return b;
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}
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2023-10-06 13:31:21 +08:00
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```
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=== "Rust"
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```rust title="climbing_stairs_dp.rs"
|
2023-10-06 14:10:18 +08:00
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/* 爬楼梯:空间优化后的动态规划 */
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fn climbing_stairs_dp_comp(n: usize) -> i32 {
|
2024-03-18 03:11:07 +08:00
|
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if n == 1 || n == 2 {
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return n as i32;
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}
|
2023-10-06 14:10:18 +08:00
|
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let (mut a, mut b) = (1, 2);
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for _ in 3..=n {
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let tmp = b;
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b = a + b;
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a = tmp;
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}
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b
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}
|
2023-10-06 13:31:21 +08:00
|
|
|
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```
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|
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=== "C"
|
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|
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```c title="climbing_stairs_dp.c"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
/* 爬楼梯:空间优化后的动态规划 */
|
|
|
|
|
int climbingStairsDPComp(int n) {
|
|
|
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if (n == 1 || n == 2)
|
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return n;
|
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|
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int a = 1, b = 2;
|
|
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for (int i = 3; i <= n; i++) {
|
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int tmp = b;
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b = a + b;
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a = tmp;
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}
|
|
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return b;
|
|
|
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}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
2024-03-25 22:43:12 +08:00
|
|
|
|
|
|
|
|
|
=== "Kotlin"
|
|
|
|
|
|
|
|
|
|
```kotlin title="climbing_stairs_dp.kt"
|
|
|
|
|
/* 爬楼梯:空间优化后的动态规划 */
|
|
|
|
|
fun climbingStairsDPComp(n: Int): Int {
|
|
|
|
|
if (n == 1 || n == 2) return n
|
|
|
|
|
var a = 1
|
|
|
|
|
var b = 2
|
|
|
|
|
for (i in 3..n) {
|
2024-04-13 21:17:44 +08:00
|
|
|
|
val temp = b
|
|
|
|
|
b += a
|
|
|
|
|
a = temp
|
2024-03-25 22:43:12 +08:00
|
|
|
|
}
|
|
|
|
|
return b
|
|
|
|
|
}
|
|
|
|
|
```
|
2024-03-31 03:53:04 +08:00
|
|
|
|
|
|
|
|
|
=== "Ruby"
|
|
|
|
|
|
|
|
|
|
```ruby title="climbing_stairs_dp.rb"
|
|
|
|
|
[class]{}-[func]{climbing_stairs_dp_comp}
|
|
|
|
|
```
|
2023-10-06 13:31:21 +08:00
|
|
|
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
|
|
|
|
```zig title="climbing_stairs_dp.zig"
|
2023-10-06 14:10:18 +08:00
|
|
|
|
// 爬楼梯:空间优化后的动态规划
|
|
|
|
|
fn climbingStairsDPComp(comptime n: usize) i32 {
|
|
|
|
|
if (n == 1 or n == 2) {
|
|
|
|
|
return @intCast(n);
|
|
|
|
|
}
|
|
|
|
|
var a: i32 = 1;
|
|
|
|
|
var b: i32 = 2;
|
|
|
|
|
for (3..n + 1) |_| {
|
|
|
|
|
var tmp = b;
|
|
|
|
|
b = a + b;
|
|
|
|
|
a = tmp;
|
|
|
|
|
}
|
|
|
|
|
return b;
|
|
|
|
|
}
|
2023-10-06 13:31:21 +08:00
|
|
|
|
```
|
|
|
|
|
|
2024-01-07 23:42:54 +08:00
|
|
|
|
??? pythontutor "可视化运行"
|
|
|
|
|
|
2024-01-09 16:00:24 +08:00
|
|
|
|
<div style="height: 477px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20climbing_stairs_dp_comp%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E7%A9%BA%E9%97%B4%E4%BC%98%E5%8C%96%E5%90%8E%E7%9A%84%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20n%0A%20%20%20%20a,%20b%20%3D%201,%202%0A%20%20%20%20for%20_%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20a,%20b%20%3D%20b,%20a%20%2B%20b%0A%20%20%20%20return%20b%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_dp_comp%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
|
|
|
|
|
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20climbing_stairs_dp_comp%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E7%A9%BA%E9%97%B4%E4%BC%98%E5%8C%96%E5%90%8E%E7%9A%84%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20n%0A%20%20%20%20a,%20b%20%3D%201,%202%0A%20%20%20%20for%20_%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20a,%20b%20%3D%20b,%20a%20%2B%20b%0A%20%20%20%20return%20b%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_dp_comp%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全屏观看 ></a></div>
|
2024-01-07 23:42:54 +08:00
|
|
|
|
|
2023-12-02 06:24:05 +08:00
|
|
|
|
观察以上代码,由于省去了数组 `dp` 占用的空间,因此空间复杂度从 $O(n)$ 降至 $O(1)$ 。
|
2023-10-06 13:31:21 +08:00
|
|
|
|
|
|
|
|
|
在动态规划问题中,当前状态往往仅与前面有限个状态有关,这时我们可以只保留必要的状态,通过“降维”来节省内存空间。**这种空间优化技巧被称为“滚动变量”或“滚动数组”**。
|