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hello-algo/docs/chapter_dynamic_programming/intro_to_dynamic_programming.md

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---
comments: true
---
# 14.1   初探动态规划
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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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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。
!!! question "爬楼梯"
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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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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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![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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<p align="center"> 图 14-1 &nbsp; 爬到第 3 阶的方案数量 </p>
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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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=== "Python"
```python title="climbing_stairs_backtrack.py"
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def backtrack(choices: list[int], state: int, n: int, res: list[int]) -> int:
"""回溯"""
# 当爬到第 n 阶时，方案数量加 1
if state == n:
res[0] += 1
# 遍历所有选择
for choice in choices:
# 剪枝：不允许越过第 n 阶
if state + choice > n:
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continue
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# 尝试：做出选择，更新状态
backtrack(choices, state + choice, n, res)
# 回退
def climbing_stairs_backtrack(n: int) -> int:
"""爬楼梯：回溯"""
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choices = [1, 2] # 可选择向上爬 1 阶或 2 阶
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state = 0 # 从第 0 阶开始爬
res = [0] # 使用 res[0] 记录方案数量
backtrack(choices, state, n, res)
return res[0]
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```
=== "C++"
```cpp title="climbing_stairs_backtrack.cpp"
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/* 回溯 */
void backtrack(vector<int> &choices, int state, int n, vector<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 (auto &choice : choices) {
// 剪枝：不允许越过第 n 阶
if (state + choice > n)
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continue;
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// 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) {
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vector<int> choices = {1, 2}; // 可选择向上爬 1 阶或 2 阶
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int state = 0; // 从第 0 阶开始爬
vector<int> res = {0}; // 使用 res[0] 记录方案数量
backtrack(choices, state, n, res);
return res[0];
}
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```
=== "Java"
```java title="climbing_stairs_backtrack.java"
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/* 回溯 */
void backtrack(List<Integer> choices, int state, int n, List<Integer> res) {
// 当爬到第 n 阶时，方案数量加 1
if (state == n)
res.set(0, res.get(0) + 1);
// 遍历所有选择
for (Integer choice : choices) {
// 剪枝：不允许越过第 n 阶
if (state + choice > n)
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continue;
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// 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) {
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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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int state = 0; // 从第 0 阶开始爬
List<Integer> res = new ArrayList<>();
res.add(0); // 使用 res[0] 记录方案数量
backtrack(choices, state, n, res);
return res.get(0);
}
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```
=== "C#"
```csharp title="climbing_stairs_backtrack.cs"
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/* 回溯 */
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void Backtrack(List<int> choices, int state, int n, List<int> res) {
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// 当爬到第 n 阶时，方案数量加 1
if (state == n)
res[0]++;
// 遍历所有选择
foreach (int choice in choices) {
// 剪枝：不允许越过第 n 阶
if (state + choice > n)
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continue;
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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
}
}
/* 爬楼梯：回溯 */
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int ClimbingStairsBacktrack(int n) {
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List<int> choices = [1, 2]; // 可选择向上爬 1 阶或 2 阶
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int state = 0; // 从第 0 阶开始爬
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List<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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```
=== "Go"
```go title="climbing_stairs_backtrack.go"
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/* 回溯 */
func backtrack(choices []int, state, n int, res []int) {
// 当爬到第 n 阶时，方案数量加 1
if state == n {
res[0] = res[0] + 1
}
// 遍历所有选择
for _, choice := range choices {
// 剪枝：不允许越过第 n 阶
if state+choice > n {
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continue
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}
// 尝试：做出选择，更新状态
backtrack(choices, state+choice, n, res)
// 回退
}
}
/* 爬楼梯：回溯 */
func climbingStairsBacktrack(n int) int {
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// 可选择向上爬 1 阶或 2 阶
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choices := []int{1, 2}
// 从第 0 阶开始爬
state := 0
res := make([]int, 1)
// 使用 res[0] 记录方案数量
res[0] = 0
backtrack(choices, state, n, res)
return res[0]
}
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```
=== "Swift"
```swift title="climbing_stairs_backtrack.swift"
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/* 回溯 */
func backtrack(choices: [Int], state: Int, n: Int, res: inout [Int]) {
// 当爬到第 n 阶时，方案数量加 1
if state == n {
res[0] += 1
}
// 遍历所有选择
for choice in choices {
// 剪枝：不允许越过第 n 阶
if state + choice > n {
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continue
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}
backtrack(choices: choices, state: state + choice, n: n, res: &res)
}
}
/* 爬楼梯：回溯 */
func climbingStairsBacktrack(n: Int) -> Int {
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let choices = [1, 2] // 可选择向上爬 1 阶或 2 阶
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let state = 0 // 从第 0 阶开始爬
var res: [Int] = []
res.append(0) // 使用 res[0] 记录方案数量
backtrack(choices: choices, state: state, n: n, res: &res)
return res[0]
}
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```
=== "JS"
```javascript title="climbing_stairs_backtrack.js"
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/* 回溯 */
function backtrack(choices, state, n, res) {
// 当爬到第 n 阶时，方案数量加 1
if (state === n) res.set(0, res.get(0) + 1);
// 遍历所有选择
for (const choice of choices) {
// 剪枝：不允许越过第 n 阶
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if (state + choice > n) continue;
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// 尝试：做出选择，更新状态
backtrack(choices, state + choice, n, res);
// 回退
}
}
/* 爬楼梯：回溯 */
function climbingStairsBacktrack(n) {
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const choices = [1, 2]; // 可选择向上爬 1 阶或 2 阶
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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);
}
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```
=== "TS"
```typescript title="climbing_stairs_backtrack.ts"
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/* 回溯 */
function backtrack(
choices: number[],
state: number,
n: number,
res: Map<0, any>
): void {
// 当爬到第 n 阶时，方案数量加 1
if (state === n) res.set(0, res.get(0) + 1);
// 遍历所有选择
for (const choice of choices) {
// 剪枝：不允许越过第 n 阶
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if (state + choice > n) continue;
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// 尝试：做出选择，更新状态
backtrack(choices, state + choice, n, res);
// 回退
}
}
/* 爬楼梯：回溯 */
function climbingStairsBacktrack(n: number): number {
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const choices = [1, 2]; // 可选择向上爬 1 阶或 2 阶
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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);
}
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```
=== "Dart"
```dart title="climbing_stairs_backtrack.dart"
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/* 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
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if (state + choice > n) continue;
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// 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) {
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List<int> choices = [1, 2]; // 可选择向上爬 1 阶或 2 阶
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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];
}
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```
=== "Rust"
```rust title="climbing_stairs_backtrack.rs"
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/* 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
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if state == n {
res[0] = res[0] + 1;
}
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// 遍历所有选择
for &choice in choices {
// 剪枝：不允许越过第 n 阶
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if state + choice > n {
continue;
}
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// 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 {
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let choices = vec![1, 2]; // 可选择向上爬 1 阶或 2 阶
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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]
}
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```
=== "C"
```c title="climbing_stairs_backtrack.c"
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/* 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)
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continue;
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// 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) {
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int choices[2] = {1, 2}; // 可选择向上爬 1 阶或 2 阶
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int state = 0; // 从第 0 阶开始爬
int *res = (int *)malloc(sizeof(int));
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*res = 0; // 使用 res[0] 记录方案数量
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int len = sizeof(choices) / sizeof(int);
backtrack(choices, state, n, res, len);
int result = *res;
free(res);
return result;
}
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```
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=== "Kotlin"
```kotlin title="climbing_stairs_backtrack.kt"
/* 回溯 */
fun backtrack(
choices: List<Int>,
state: Int,
n: Int,
res: MutableList<Int>
) {
// 当爬到第 n 阶时，方案数量加 1
if (state == n) res[0] = res[0] + 1
// 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
val res = ArrayList<Int>()
res.add(0) // 使用 res[0] 记录方案数量
backtrack(choices, state, n, res)
return res[0]
}
```
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=== "Ruby"
```ruby title="climbing_stairs_backtrack.rb"
[class]{}-[func]{backtrack}
[class]{}-[func]{climbing_stairs_backtrack}
```
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=== "Zig"
```zig title="climbing_stairs_backtrack.zig"
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// 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) {
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continue;
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}
// 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 {
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var choices = [_]i32{ 1, 2 }; // 可选择向上爬 1 阶或 2 阶
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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];
}
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```
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??? pythontutor "可视化运行"
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<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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## 14.1.1 &nbsp; 方法一：暴力搜索
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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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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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$$
dp[i-1], dp[i-2], \dots, dp[2], dp[1]
$$
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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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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：
$$
dp[i] = dp[i-1] + dp[i-2]
$$
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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![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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<p align="center"> 图 14-2 &nbsp; 方案数量递推关系 </p>
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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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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=== "Python"
```python title="climbing_stairs_dfs.py"
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def dfs(i: int) -> int:
"""搜索"""
# 已知 dp[1] 和 dp[2] ，返回之
if i == 1 or i == 2:
return i
# dp[i] = dp[i-1] + dp[i-2]
count = dfs(i - 1) + dfs(i - 2)
return count
def climbing_stairs_dfs(n: int) -> int:
"""爬楼梯：搜索"""
return dfs(n)
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```
=== "C++"
```cpp title="climbing_stairs_dfs.cpp"
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/* 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);
}
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```
=== "Java"
```java title="climbing_stairs_dfs.java"
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/* 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);
}
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```
=== "C#"
```csharp title="climbing_stairs_dfs.cs"
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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
if (i == 1 || i == 2)
return i;
// 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;
}
/* 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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```
=== "Go"
```go title="climbing_stairs_dfs.go"
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/* 搜索 */
func 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]
count := dfs(i-1) + dfs(i-2)
return count
}
/* 爬楼梯：搜索 */
func climbingStairsDFS(n int) int {
return dfs(n)
}
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```
=== "Swift"
```swift title="climbing_stairs_dfs.swift"
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/* 搜索 */
func 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]
let count = dfs(i: i - 1) + dfs(i: i - 2)
return count
}
/* 爬楼梯：搜索 */
func climbingStairsDFS(n: Int) -> Int {
dfs(i: n)
}
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```
=== "JS"
```javascript title="climbing_stairs_dfs.js"
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/* 搜索 */
function dfs(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]
const count = dfs(i - 1) + dfs(i - 2);
return count;
}
/* 爬楼梯：搜索 */
function climbingStairsDFS(n) {
return dfs(n);
}
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```
=== "TS"
```typescript title="climbing_stairs_dfs.ts"
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/* 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);
}
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```
=== "Dart"
```dart title="climbing_stairs_dfs.dart"
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/* 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);
}
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```
=== "Rust"
```rust title="climbing_stairs_dfs.rs"
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/* 搜索 */
fn dfs(i: usize) -> i32 {
// 已知 dp[1] 和 dp[2] ，返回之
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if i == 1 || i == 2 {
return i as i32;
}
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// 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 {
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dfs(n)
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}
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```
=== "C"
```c title="climbing_stairs_dfs.c"
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/* 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);
}
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```
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=== "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)
}
```
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=== "Ruby"
```ruby title="climbing_stairs_dfs.rb"
[class]{}-[func]{dfs}
[class]{}-[func]{climbing_stairs_dfs}
```
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=== "Zig"
```zig title="climbing_stairs_dfs.zig"
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// 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);
}
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```
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??? pythontutor "可视化运行"
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<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>
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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。
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![爬楼梯对应递归树](intro_to_dynamic_programming.assets/climbing_stairs_dfs_tree.png){ class="animation-figure" }
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<p align="center"> 图 14-3 &nbsp; 爬楼梯对应递归树 </p>
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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 $dp[7]$ 。
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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。
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## 14.1.2 &nbsp; 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。
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代码如下所示：
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=== "Python"
```python title="climbing_stairs_dfs_mem.py"
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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)
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```
=== "C++"
```cpp title="climbing_stairs_dfs_mem.cpp"
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/* 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);
}
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```
=== "Java"
```java title="climbing_stairs_dfs_mem.java"
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/* 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);
}
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```
=== "C#"
```csharp title="climbing_stairs_dfs_mem.cs"
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/* 记忆化搜索 */
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int DFS(int i, int[] mem) {
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// 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]
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int count = DFS(i - 1, mem) + DFS(i - 2, mem);
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// 记录 dp[i]
mem[i] = count;
return count;
}
/* 爬楼梯：记忆化搜索 */
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int ClimbingStairsDFSMem(int n) {
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// 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);
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return DFS(n, mem);
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}
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```
=== "Go"
```go title="climbing_stairs_dfs_mem.go"
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/* 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)
}
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```
=== "Swift"
```swift title="climbing_stairs_dfs_mem.swift"
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/* 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)
}
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```
=== "JS"
```javascript title="climbing_stairs_dfs_mem.js"
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/* 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);
}
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```
=== "TS"
```typescript title="climbing_stairs_dfs_mem.ts"
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/* 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);
}
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```
=== "Dart"
```dart title="climbing_stairs_dfs_mem.dart"
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/* 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);
}
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```
=== "Rust"
```rust title="climbing_stairs_dfs_mem.rs"
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/* 记忆化搜索 */
fn dfs(i: usize, mem: &mut [i32]) -> i32 {
// 已知 dp[1] 和 dp[2] ，返回之
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if i == 1 || i == 2 {
return i as i32;
}
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// 若存在记录 dp[i] ，则直接返回之
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if mem[i] != -1 {
return mem[i];
}
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// 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)
}
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```
=== "C"
```c title="climbing_stairs_dfs_mem.c"
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/* 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;
}
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```
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=== "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)
Arrays.fill(mem, -1)
return dfs(n, mem)
}
```
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=== "Ruby"
```ruby title="climbing_stairs_dfs_mem.rb"
[class]{}-[func]{dfs}
[class]{}-[func]{climbing_stairs_dfs_mem}
```
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=== "Zig"
```zig title="climbing_stairs_dfs_mem.zig"
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// 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);
}
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```
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??? pythontutor "可视化运行"
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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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![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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<p align="center"> 图 14-4 &nbsp; 记忆化搜索对应递归树 </p>
## 14.1.3 &nbsp; 方法三：动态规划
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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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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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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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=== "Python"
```python title="climbing_stairs_dp.py"
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def climbing_stairs_dp(n: int) -> int:
"""爬楼梯：动态规划"""
if n == 1 or n == 2:
return n
# 初始化 dp 表，用于存储子问题的解
dp = [0] * (n + 1)
# 初始状态：预设最小子问题的解
dp[1], dp[2] = 1, 2
# 状态转移：从较小子问题逐步求解较大子问题
for i in range(3, n + 1):
dp[i] = dp[i - 1] + dp[i - 2]
return dp[n]
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```
=== "C++"
```cpp title="climbing_stairs_dp.cpp"
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/* 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
vector<int> dp(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 (int i = 3; i <= n; i++) {
dp[i] = dp[i - 1] + dp[i - 2];
}
return dp[n];
}
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```
=== "Java"
```java title="climbing_stairs_dp.java"
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/* 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 = new int[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 (int i = 3; i <= n; i++) {
dp[i] = dp[i - 1] + dp[i - 2];
}
return dp[n];
}
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```
=== "C#"
```csharp title="climbing_stairs_dp.cs"
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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)
return n;
// 初始化 dp 表，用于存储子问题的解
int[] dp = new int[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 (int i = 3; i <= n; i++) {
dp[i] = dp[i - 1] + dp[i - 2];
}
return dp[n];
}
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```
=== "Go"
```go title="climbing_stairs_dp.go"
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/* 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
dp := make([]int, 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 := 3; i <= n; i++ {
dp[i] = dp[i-1] + dp[i-2]
}
return dp[n]
}
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```
=== "Swift"
```swift title="climbing_stairs_dp.swift"
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/* 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
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for i in 3 ... n {
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dp[i] = dp[i - 1] + dp[i - 2]
}
return dp[n]
}
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```
=== "JS"
```javascript title="climbing_stairs_dp.js"
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/* 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];
}
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```
=== "TS"
```typescript title="climbing_stairs_dp.ts"
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/* 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];
}
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```
=== "Dart"
```dart title="climbing_stairs_dp.dart"
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/* 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];
}
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```
=== "Rust"
```rust title="climbing_stairs_dp.rs"
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/* 爬楼梯：动态规划 */
fn climbing_stairs_dp(n: usize) -> i32 {
// 已知 dp[1] 和 dp[2] ，返回之
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if n == 1 || n == 2 {
return n as i32;
}
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// 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]
}
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```
=== "C"
```c title="climbing_stairs_dp.c"
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/* 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;
}
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```
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=== "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]
}
```
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=== "Ruby"
```ruby title="climbing_stairs_dp.rb"
[class]{}-[func]{climbing_stairs_dp}
```
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=== "Zig"
```zig title="climbing_stairs_dp.zig"
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// 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];
}
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```
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??? pythontutor "可视化运行"
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<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>
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图 14-5 模拟了以上代码的执行过程。
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![爬楼梯的动态规划过程](intro_to_dynamic_programming.assets/climbing_stairs_dp.png){ class="animation-figure" }
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<p align="center"> 图 14-5 &nbsp; 爬楼梯的动态规划过程 </p>
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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 $i$ 。
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根据以上内容，我们可以总结出动态规划的常用术语。
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- 将数组 `dp` 称为<u>dp 表</u>，$dp[i]$ 表示状态 $i$ 对应子问题的解。
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- 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>。
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## 14.1.4 &nbsp; 空间优化
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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：
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=== "Python"
```python title="climbing_stairs_dp.py"
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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
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```
=== "C++"
```cpp title="climbing_stairs_dp.cpp"
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/* 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;
}
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```
=== "Java"
```java title="climbing_stairs_dp.java"
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/* 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;
}
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```
=== "C#"
```csharp title="climbing_stairs_dp.cs"
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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;
int a = 1, b = 2;
for (int i = 3; i <= n; i++) {
int tmp = b;
b = a + b;
a = tmp;
}
return b;
}
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```
=== "Go"
```go title="climbing_stairs_dp.go"
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/* 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
}
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```
=== "Swift"
```swift title="climbing_stairs_dp.swift"
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/* 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
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for _ in 3 ... n {
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(a, b) = (b, a + b)
}
return b
}
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```
=== "JS"
```javascript title="climbing_stairs_dp.js"
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/* 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;
}
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```
=== "TS"
```typescript title="climbing_stairs_dp.ts"
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/* 爬楼梯：空间优化后的动态规划 */
function climbingStairsDPComp(n: number): number {
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;
}
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```
=== "Dart"
```dart title="climbing_stairs_dp.dart"
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/* 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;
}
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```
=== "Rust"
```rust title="climbing_stairs_dp.rs"
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/* 爬楼梯：空间优化后的动态规划 */
fn climbing_stairs_dp_comp(n: usize) -> i32 {
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if n == 1 || n == 2 {
return n as i32;
}
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let (mut a, mut b) = (1, 2);
for _ in 3..=n {
let tmp = b;
b = a + b;
a = tmp;
}
b
}
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```
=== "C"
```c title="climbing_stairs_dp.c"
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/* 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;
}
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```
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=== "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) {
val tmp = b
b += a
a = tmp
}
return b
}
```
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=== "Ruby"
```ruby title="climbing_stairs_dp.rb"
[class]{}-[func]{climbing_stairs_dp_comp}
```
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=== "Zig"
```zig title="climbing_stairs_dp.zig"
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// 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;
}
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```
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??? pythontutor "可视化运行"
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<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>
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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 $O(1)$ 。
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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”**。
Reference in a new issue Copy permalink