hello-algo/zh-Hant/docs/chapter_dynamic_programming/intro_to_dynamic_programming.md

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---
comments: true
---
# 14.1   初探動態規劃
<u>動態規劃dynamic programming</u>是一個重要的演算法範式,它將一個問題分解為一系列更小的子問題,並透過儲存子問題的解來避免重複計算,從而大幅提升時間效率。
在本節中,我們從一個經典例題入手,先給出它的暴力回溯解法,觀察其中包含的重疊子問題,再逐步導出更高效的動態規劃解法。
!!! question "爬樓梯"
給定一個共有 $n$ 階的樓梯,你每步可以上 $1$ 階或者 $2$ 階,請問有多少種方案可以爬到樓頂?
如圖 14-1 所示,對於一個 $3$ 階樓梯,共有 $3$ 種方案可以爬到樓頂。
![爬到第 3 階的方案數量](intro_to_dynamic_programming.assets/climbing_stairs_example.png){ class="animation-figure" }
<p align="center"> 圖 14-1 &nbsp; 爬到第 3 階的方案數量 </p>
本題的目標是求解方案數量,**我們可以考慮透過回溯來窮舉所有可能性**。具體來說,將爬樓梯想象為一個多輪選擇的過程:從地面出發,每輪選擇上 $1$ 階或 $2$ 階,每當到達樓梯頂部時就將方案數量加 $1$ ,當越過樓梯頂部時就將其剪枝。程式碼如下所示:
=== "Python"
```python title="climbing_stairs_backtrack.py"
def backtrack(choices: list[int], state: int, n: int, res: list[int]) -> int:
"""回溯"""
# 當爬到第 n 階時,方案數量加 1
if state == n:
res[0] += 1
# 走訪所有選擇
for choice in choices:
# 剪枝:不允許越過第 n 階
if state + choice > n:
continue
# 嘗試:做出選擇,更新狀態
backtrack(choices, state + choice, n, res)
# 回退
def climbing_stairs_backtrack(n: int) -> int:
"""爬樓梯:回溯"""
choices = [1, 2] # 可選擇向上爬 1 階或 2 階
state = 0 # 從第 0 階開始爬
res = [0] # 使用 res[0] 記錄方案數量
backtrack(choices, state, n, res)
return res[0]
```
=== "C++"
```cpp title="climbing_stairs_backtrack.cpp"
/* 回溯 */
void backtrack(vector<int> &choices, int state, int n, vector<int> &res) {
// 當爬到第 n 階時,方案數量加 1
if (state == n)
res[0]++;
// 走訪所有選擇
for (auto &choice : choices) {
// 剪枝:不允許越過第 n 階
if (state + choice > n)
continue;
// 嘗試:做出選擇,更新狀態
backtrack(choices, state + choice, n, res);
// 回退
}
}
/* 爬樓梯:回溯 */
int climbingStairsBacktrack(int n) {
vector<int> choices = {1, 2}; // 可選擇向上爬 1 階或 2 階
int state = 0; // 從第 0 階開始爬
vector<int> res = {0}; // 使用 res[0] 記錄方案數量
backtrack(choices, state, n, res);
return res[0];
}
```
=== "Java"
```java title="climbing_stairs_backtrack.java"
/* 回溯 */
void backtrack(List<Integer> choices, int state, int n, List<Integer> res) {
// 當爬到第 n 階時,方案數量加 1
if (state == n)
res.set(0, res.get(0) + 1);
// 走訪所有選擇
for (Integer choice : choices) {
// 剪枝:不允許越過第 n 階
if (state + choice > n)
continue;
// 嘗試:做出選擇,更新狀態
backtrack(choices, state + choice, n, res);
// 回退
}
}
/* 爬樓梯:回溯 */
int climbingStairsBacktrack(int n) {
List<Integer> choices = Arrays.asList(1, 2); // 可選擇向上爬 1 階或 2 階
int state = 0; // 從第 0 階開始爬
List<Integer> res = new ArrayList<>();
res.add(0); // 使用 res[0] 記錄方案數量
backtrack(choices, state, n, res);
return res.get(0);
}
```
=== "C#"
```csharp title="climbing_stairs_backtrack.cs"
/* 回溯 */
void Backtrack(List<int> choices, int state, int n, List<int> res) {
// 當爬到第 n 階時,方案數量加 1
if (state == n)
res[0]++;
// 走訪所有選擇
foreach (int choice in choices) {
// 剪枝:不允許越過第 n 階
if (state + choice > n)
continue;
// 嘗試:做出選擇,更新狀態
Backtrack(choices, state + choice, n, res);
// 回退
}
}
/* 爬樓梯:回溯 */
int ClimbingStairsBacktrack(int n) {
List<int> choices = [1, 2]; // 可選擇向上爬 1 階或 2 階
int state = 0; // 從第 0 階開始爬
List<int> res = [0]; // 使用 res[0] 記錄方案數量
Backtrack(choices, state, n, res);
return res[0];
}
```
=== "Go"
```go title="climbing_stairs_backtrack.go"
/* 回溯 */
func backtrack(choices []int, state, n int, res []int) {
// 當爬到第 n 階時,方案數量加 1
if state == n {
res[0] = res[0] + 1
}
// 走訪所有選擇
for _, choice := range choices {
// 剪枝:不允許越過第 n 階
if state+choice > n {
continue
}
// 嘗試:做出選擇,更新狀態
backtrack(choices, state+choice, n, res)
// 回退
}
}
/* 爬樓梯:回溯 */
func climbingStairsBacktrack(n int) int {
// 可選擇向上爬 1 階或 2 階
choices := []int{1, 2}
// 從第 0 階開始爬
state := 0
res := make([]int, 1)
// 使用 res[0] 記錄方案數量
res[0] = 0
backtrack(choices, state, n, res)
return res[0]
}
```
=== "Swift"
```swift title="climbing_stairs_backtrack.swift"
/* 回溯 */
func backtrack(choices: [Int], state: Int, n: Int, res: inout [Int]) {
// 當爬到第 n 階時,方案數量加 1
if state == n {
res[0] += 1
}
// 走訪所有選擇
for choice in choices {
// 剪枝:不允許越過第 n 階
if state + choice > n {
continue
}
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// 嘗試:做出選擇,更新狀態
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backtrack(choices: choices, state: state + choice, n: n, res: &res)
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// 回退
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}
}
/* 爬樓梯:回溯 */
func climbingStairsBacktrack(n: Int) -> Int {
let choices = [1, 2] // 可選擇向上爬 1 階或 2 階
let state = 0 // 從第 0 階開始爬
var res: [Int] = []
res.append(0) // 使用 res[0] 記錄方案數量
backtrack(choices: choices, state: state, n: n, res: &res)
return res[0]
}
```
=== "JS"
```javascript title="climbing_stairs_backtrack.js"
/* 回溯 */
function backtrack(choices, state, n, res) {
// 當爬到第 n 階時,方案數量加 1
if (state === n) res.set(0, res.get(0) + 1);
// 走訪所有選擇
for (const choice of choices) {
// 剪枝:不允許越過第 n 階
if (state + choice > n) continue;
// 嘗試:做出選擇,更新狀態
backtrack(choices, state + choice, n, res);
// 回退
}
}
/* 爬樓梯:回溯 */
function climbingStairsBacktrack(n) {
const choices = [1, 2]; // 可選擇向上爬 1 階或 2 階
const state = 0; // 從第 0 階開始爬
const res = new Map();
res.set(0, 0); // 使用 res[0] 記錄方案數量
backtrack(choices, state, n, res);
return res.get(0);
}
```
=== "TS"
```typescript title="climbing_stairs_backtrack.ts"
/* 回溯 */
function backtrack(
choices: number[],
state: number,
n: number,
res: Map<0, any>
): void {
// 當爬到第 n 階時,方案數量加 1
if (state === n) res.set(0, res.get(0) + 1);
// 走訪所有選擇
for (const choice of choices) {
// 剪枝:不允許越過第 n 階
if (state + choice > n) continue;
// 嘗試:做出選擇,更新狀態
backtrack(choices, state + choice, n, res);
// 回退
}
}
/* 爬樓梯:回溯 */
function climbingStairsBacktrack(n: number): number {
const choices = [1, 2]; // 可選擇向上爬 1 階或 2 階
const state = 0; // 從第 0 階開始爬
const res = new Map();
res.set(0, 0); // 使用 res[0] 記錄方案數量
backtrack(choices, state, n, res);
return res.get(0);
}
```
=== "Dart"
```dart title="climbing_stairs_backtrack.dart"
/* 回溯 */
void backtrack(List<int> choices, int state, int n, List<int> res) {
// 當爬到第 n 階時,方案數量加 1
if (state == n) {
res[0]++;
}
// 走訪所有選擇
for (int choice in choices) {
// 剪枝:不允許越過第 n 階
if (state + choice > n) continue;
// 嘗試:做出選擇,更新狀態
backtrack(choices, state + choice, n, res);
// 回退
}
}
/* 爬樓梯:回溯 */
int climbingStairsBacktrack(int n) {
List<int> choices = [1, 2]; // 可選擇向上爬 1 階或 2 階
int state = 0; // 從第 0 階開始爬
List<int> res = [];
res.add(0); // 使用 res[0] 記錄方案數量
backtrack(choices, state, n, res);
return res[0];
}
```
=== "Rust"
```rust title="climbing_stairs_backtrack.rs"
/* 回溯 */
fn backtrack(choices: &[i32], state: i32, n: i32, res: &mut [i32]) {
// 當爬到第 n 階時,方案數量加 1
if state == n {
res[0] = res[0] + 1;
}
// 走訪所有選擇
for &choice in choices {
// 剪枝:不允許越過第 n 階
if state + choice > n {
continue;
}
// 嘗試:做出選擇,更新狀態
backtrack(choices, state + choice, n, res);
// 回退
}
}
/* 爬樓梯:回溯 */
fn climbing_stairs_backtrack(n: usize) -> i32 {
let choices = vec![1, 2]; // 可選擇向上爬 1 階或 2 階
let state = 0; // 從第 0 階開始爬
let mut res = Vec::new();
res.push(0); // 使用 res[0] 記錄方案數量
backtrack(&choices, state, n as i32, &mut res);
res[0]
}
```
=== "C"
```c title="climbing_stairs_backtrack.c"
/* 回溯 */
void backtrack(int *choices, int state, int n, int *res, int len) {
// 當爬到第 n 階時,方案數量加 1
if (state == n)
res[0]++;
// 走訪所有選擇
for (int i = 0; i < len; i++) {
int choice = choices[i];
// 剪枝:不允許越過第 n 階
if (state + choice > n)
continue;
// 嘗試:做出選擇,更新狀態
backtrack(choices, state + choice, n, res, len);
// 回退
}
}
/* 爬樓梯:回溯 */
int climbingStairsBacktrack(int n) {
int choices[2] = {1, 2}; // 可選擇向上爬 1 階或 2 階
int state = 0; // 從第 0 階開始爬
int *res = (int *)malloc(sizeof(int));
*res = 0; // 使用 res[0] 記錄方案數量
int len = sizeof(choices) / sizeof(int);
backtrack(choices, state, n, res, len);
int result = *res;
free(res);
return result;
}
```
=== "Kotlin"
```kotlin title="climbing_stairs_backtrack.kt"
/* 回溯 */
fun backtrack(
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choices: MutableList<Int>,
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state: Int,
n: Int,
res: MutableList<Int>
) {
// 當爬到第 n 階時,方案數量加 1
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if (state == n)
res[0] = res[0] + 1
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// 走訪所有選擇
for (choice in choices) {
// 剪枝:不允許越過第 n 階
if (state + choice > n) continue
// 嘗試:做出選擇,更新狀態
backtrack(choices, state + choice, n, res)
// 回退
}
}
/* 爬樓梯:回溯 */
fun climbingStairsBacktrack(n: Int): Int {
val choices = mutableListOf(1, 2) // 可選擇向上爬 1 階或 2 階
val state = 0 // 從第 0 階開始爬
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val res = mutableListOf<Int>()
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res.add(0) // 使用 res[0] 記錄方案數量
backtrack(choices, state, n, res)
return res[0]
}
```
=== "Ruby"
```ruby title="climbing_stairs_backtrack.rb"
[class]{}-[func]{backtrack}
[class]{}-[func]{climbing_stairs_backtrack}
```
=== "Zig"
```zig title="climbing_stairs_backtrack.zig"
// 回溯
fn backtrack(choices: []i32, state: i32, n: i32, res: std.ArrayList(i32)) void {
// 當爬到第 n 階時,方案數量加 1
if (state == n) {
res.items[0] = res.items[0] + 1;
}
// 走訪所有選擇
for (choices) |choice| {
// 剪枝:不允許越過第 n 階
if (state + choice > n) {
continue;
}
// 嘗試:做出選擇,更新狀態
backtrack(choices, state + choice, n, res);
// 回退
}
}
// 爬樓梯:回溯
fn climbingStairsBacktrack(n: usize) !i32 {
var choices = [_]i32{ 1, 2 }; // 可選擇向上爬 1 階或 2 階
var state: i32 = 0; // 從第 0 階開始爬
var res = std.ArrayList(i32).init(std.heap.page_allocator);
defer res.deinit();
try res.append(0); // 使用 res[0] 記錄方案數量
backtrack(&choices, state, @intCast(n), res);
return res.items[0];
}
```
??? pythontutor "視覺化執行"
2024-04-11 01:11:20 +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%2C%20state%3A%20int%2C%20n%3A%20int%2C%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%E7%95%B6%E7%88%AC%E5%88%B0%E7%AC%AC%20n%20%E9%9A%8E%E6%99%82%EF%BC%8C%E6%96%B9%E6%A1%88%E6%95%B8%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%E8%B5%B0%E8%A8%AA%E6%89%80%E6%9C%89%E9%81%B8%E6%93%87%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%A8%B1%E8%B6%8A%E9%81%8E%E7%AC%AC%20n%20%E9%9A%8E%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%98%97%E8%A9%A6%EF%BC%9A%E5%81%9A%E5%87%BA%E9%81%B8%E6%93%87%EF%BC%8C%E6%9B%B4%E6%96%B0%E7%8B%80%E6%85%8B%0A%20%20%20%20%20%20%20%20backtrack%28choices%2C%20state%20%2B%20choice%2C%20n%2C%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%A8%93%E6%A2%AF%EF%BC%9A%E5%9B%9E%E6%BA%AF%22%22%22%0A%20%20%20%20choices%20%3D%20%5B1%2C%202%5D%20%20%23%20%E5%8F%AF%E9%81%B8%E6%93%87%E5%90%91%E4%B8%8A%E7%88%AC%201%20%E9%9A%8E%E6%88%96%202%20%E9%9A%8E%0A%20%20%20%20state%20%3D%200%20%20%23%20%E5%BE%9E%E7%AC%AC%200%20%E9%9A%8E%E9%96%8B%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%A8%98%E9%8C%84%E6%96%B9%E6%A1%88%E6%95%B8%E9%87%8F%0A%20%20%20%20backtrack%28choices%2C%20state%2C%20n%2C%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%9A%8E%E6%A8%93%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A8%AE%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>
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2024-04-06 03:02:20 +08:00
## 14.1.1 &nbsp; 方法一:暴力搜尋
回溯演算法通常並不顯式地對問題進行拆解,而是將求解問題看作一系列決策步驟,透過試探和剪枝,搜尋所有可能的解。
我們可以嘗試從問題分解的角度分析這道題。設爬到第 $i$ 階共有 $dp[i]$ 種方案,那麼 $dp[i]$ 就是原問題,其子問題包括:
$$
dp[i-1], dp[i-2], \dots, dp[2], dp[1]
$$
由於每輪只能上 $1$ 階或 $2$ 階,因此當我們站在第 $i$ 階樓梯上時,上一輪只可能站在第 $i - 1$ 階或第 $i - 2$ 階上。換句話說,我們只能從第 $i -1$ 階或第 $i - 2$ 階邁向第 $i$ 階。
由此便可得出一個重要推論:**爬到第 $i - 1$ 階的方案數加上爬到第 $i - 2$ 階的方案數就等於爬到第 $i$ 階的方案數**。公式如下:
$$
dp[i] = dp[i-1] + dp[i-2]
$$
這意味著在爬樓梯問題中,各個子問題之間存在遞推關係,**原問題的解可以由子問題的解構建得來**。圖 14-2 展示了該遞推關係。
![方案數量遞推關係](intro_to_dynamic_programming.assets/climbing_stairs_state_transfer.png){ class="animation-figure" }
<p align="center"> 圖 14-2 &nbsp; 方案數量遞推關係 </p>
我們可以根據遞推公式得到暴力搜尋解法。以 $dp[n]$ 為起始點,**遞迴地將一個較大問題拆解為兩個較小問題的和**,直至到達最小子問題 $dp[1]$ 和 $dp[2]$ 時返回。其中,最小子問題的解是已知的,即 $dp[1] = 1$、$dp[2] = 2$ ,表示爬到第 $1$、$2$ 階分別有 $1$、$2$ 種方案。
觀察以下程式碼,它和標準回溯程式碼都屬於深度優先搜尋,但更加簡潔:
=== "Python"
```python title="climbing_stairs_dfs.py"
def dfs(i: int) -> int:
"""搜尋"""
# 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:
"""爬樓梯:搜尋"""
return dfs(n)
```
=== "C++"
```cpp title="climbing_stairs_dfs.cpp"
/* 搜尋 */
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;
}
/* 爬樓梯:搜尋 */
int climbingStairsDFS(int n) {
return dfs(n);
}
```
=== "Java"
```java title="climbing_stairs_dfs.java"
/* 搜尋 */
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;
}
/* 爬樓梯:搜尋 */
int climbingStairsDFS(int n) {
return dfs(n);
}
```
=== "C#"
```csharp title="climbing_stairs_dfs.cs"
/* 搜尋 */
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;
}
/* 爬樓梯:搜尋 */
int ClimbingStairsDFS(int n) {
return DFS(n);
}
```
=== "Go"
```go title="climbing_stairs_dfs.go"
/* 搜尋 */
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
}
/* 爬樓梯:搜尋 */
func climbingStairsDFS(n int) int {
return dfs(n)
}
```
=== "Swift"
```swift title="climbing_stairs_dfs.swift"
/* 搜尋 */
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
}
/* 爬樓梯:搜尋 */
func climbingStairsDFS(n: Int) -> Int {
dfs(i: n)
}
```
=== "JS"
```javascript title="climbing_stairs_dfs.js"
/* 搜尋 */
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;
}
/* 爬樓梯:搜尋 */
function climbingStairsDFS(n) {
return dfs(n);
}
```
=== "TS"
```typescript title="climbing_stairs_dfs.ts"
/* 搜尋 */
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;
}
/* 爬樓梯:搜尋 */
function climbingStairsDFS(n: number): number {
return dfs(n);
}
```
=== "Dart"
```dart title="climbing_stairs_dfs.dart"
/* 搜尋 */
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;
}
/* 爬樓梯:搜尋 */
int climbingStairsDFS(int n) {
return dfs(n);
}
```
=== "Rust"
```rust title="climbing_stairs_dfs.rs"
/* 搜尋 */
fn dfs(i: usize) -> i32 {
// 已知 dp[1] 和 dp[2] ,返回之
if i == 1 || i == 2 {
return i as i32;
}
// dp[i] = dp[i-1] + dp[i-2]
let count = dfs(i - 1) + dfs(i - 2);
count
}
/* 爬樓梯:搜尋 */
fn climbing_stairs_dfs(n: usize) -> i32 {
dfs(n)
}
```
=== "C"
```c title="climbing_stairs_dfs.c"
/* 搜尋 */
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;
}
/* 爬樓梯:搜尋 */
int climbingStairsDFS(int n) {
return dfs(n);
}
```
=== "Kotlin"
```kotlin title="climbing_stairs_dfs.kt"
/* 搜尋 */
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
}
/* 爬樓梯:搜尋 */
fun climbingStairsDFS(n: Int): Int {
return dfs(n)
}
```
=== "Ruby"
```ruby title="climbing_stairs_dfs.rb"
[class]{}-[func]{dfs}
[class]{}-[func]{climbing_stairs_dfs}
```
=== "Zig"
```zig title="climbing_stairs_dfs.zig"
// 搜尋
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;
}
// 爬樓梯:搜尋
fn climbingStairsDFS(comptime n: usize) i32 {
return dfs(n);
}
```
??? pythontutor "視覺化執行"
2024-04-11 01:11:20 +08:00
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2024-04-06 03:02:20 +08:00
圖 14-3 展示了暴力搜尋形成的遞迴樹。對於問題 $dp[n]$ ,其遞迴樹的深度為 $n$ ,時間複雜度為 $O(2^n)$ 。指數階屬於爆炸式增長,如果我們輸入一個比較大的 $n$ ,則會陷入漫長的等待之中。
![爬樓梯對應遞迴樹](intro_to_dynamic_programming.assets/climbing_stairs_dfs_tree.png){ class="animation-figure" }
<p align="center"> 圖 14-3 &nbsp; 爬樓梯對應遞迴樹 </p>
觀察圖 14-3 **指數階的時間複雜度是“重疊子問題”導致的**。例如 $dp[9]$ 被分解為 $dp[8]$ 和 $dp[7]$ $dp[8]$ 被分解為 $dp[7]$ 和 $dp[6]$ ,兩者都包含子問題 $dp[7]$ 。
以此類推,子問題中包含更小的重疊子問題,子子孫孫無窮盡也。絕大部分計算資源都浪費在這些重疊的子問題上。
## 14.1.2 &nbsp; 方法二:記憶化搜尋
為了提升演算法效率,**我們希望所有的重疊子問題都只被計算一次**。為此,我們宣告一個陣列 `mem` 來記錄每個子問題的解,並在搜尋過程中將重疊子問題剪枝。
1. 當首次計算 $dp[i]$ 時,我們將其記錄至 `mem[i]` ,以便之後使用。
2. 當再次需要計算 $dp[i]$ 時,我們便可直接從 `mem[i]` 中獲取結果,從而避免重複計算該子問題。
程式碼如下所示:
=== "Python"
```python title="climbing_stairs_dfs_mem.py"
def dfs(i: int, mem: list[int]) -> int:
"""記憶化搜尋"""
# 已知 dp[1] 和 dp[2] ,返回之
if i == 1 or i == 2:
return i
# 若存在記錄 dp[i] ,則直接返回之
if mem[i] != -1:
return mem[i]
# dp[i] = dp[i-1] + dp[i-2]
count = dfs(i - 1, mem) + dfs(i - 2, mem)
# 記錄 dp[i]
mem[i] = count
return count
def climbing_stairs_dfs_mem(n: int) -> int:
"""爬樓梯:記憶化搜尋"""
# mem[i] 記錄爬到第 i 階的方案總數,-1 代表無記錄
mem = [-1] * (n + 1)
return dfs(n, mem)
```
=== "C++"
```cpp title="climbing_stairs_dfs_mem.cpp"
/* 記憶化搜尋 */
int dfs(int i, vector<int> &mem) {
// 已知 dp[1] 和 dp[2] ,返回之
if (i == 1 || i == 2)
return i;
// 若存在記錄 dp[i] ,則直接返回之
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);
// 記錄 dp[i]
mem[i] = count;
return count;
}
/* 爬樓梯:記憶化搜尋 */
int climbingStairsDFSMem(int n) {
// mem[i] 記錄爬到第 i 階的方案總數,-1 代表無記錄
vector<int> mem(n + 1, -1);
return dfs(n, mem);
}
```
=== "Java"
```java title="climbing_stairs_dfs_mem.java"
/* 記憶化搜尋 */
int dfs(int i, int[] mem) {
// 已知 dp[1] 和 dp[2] ,返回之
if (i == 1 || i == 2)
return i;
// 若存在記錄 dp[i] ,則直接返回之
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);
// 記錄 dp[i]
mem[i] = count;
return count;
}
/* 爬樓梯:記憶化搜尋 */
int climbingStairsDFSMem(int n) {
// mem[i] 記錄爬到第 i 階的方案總數,-1 代表無記錄
int[] mem = new int[n + 1];
Arrays.fill(mem, -1);
return dfs(n, mem);
}
```
=== "C#"
```csharp title="climbing_stairs_dfs_mem.cs"
/* 記憶化搜尋 */
int DFS(int i, int[] mem) {
// 已知 dp[1] 和 dp[2] ,返回之
if (i == 1 || i == 2)
return i;
// 若存在記錄 dp[i] ,則直接返回之
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);
// 記錄 dp[i]
mem[i] = count;
return count;
}
/* 爬樓梯:記憶化搜尋 */
int ClimbingStairsDFSMem(int n) {
// mem[i] 記錄爬到第 i 階的方案總數,-1 代表無記錄
int[] mem = new int[n + 1];
Array.Fill(mem, -1);
return DFS(n, mem);
}
```
=== "Go"
```go title="climbing_stairs_dfs_mem.go"
/* 記憶化搜尋 */
func dfsMem(i int, mem []int) int {
// 已知 dp[1] 和 dp[2] ,返回之
if i == 1 || i == 2 {
return i
}
// 若存在記錄 dp[i] ,則直接返回之
if mem[i] != -1 {
return mem[i]
}
// dp[i] = dp[i-1] + dp[i-2]
count := dfsMem(i-1, mem) + dfsMem(i-2, mem)
// 記錄 dp[i]
mem[i] = count
return count
}
/* 爬樓梯:記憶化搜尋 */
func climbingStairsDFSMem(n int) int {
// mem[i] 記錄爬到第 i 階的方案總數,-1 代表無記錄
mem := make([]int, n+1)
for i := range mem {
mem[i] = -1
}
return dfsMem(n, mem)
}
```
=== "Swift"
```swift title="climbing_stairs_dfs_mem.swift"
/* 記憶化搜尋 */
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
}
// 若存在記錄 dp[i] ,則直接返回之
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)
// 記錄 dp[i]
mem[i] = count
return count
}
/* 爬樓梯:記憶化搜尋 */
func climbingStairsDFSMem(n: Int) -> Int {
// mem[i] 記錄爬到第 i 階的方案總數,-1 代表無記錄
var mem = Array(repeating: -1, count: n + 1)
return dfs(i: n, mem: &mem)
}
```
=== "JS"
```javascript title="climbing_stairs_dfs_mem.js"
/* 記憶化搜尋 */
function dfs(i, mem) {
// 已知 dp[1] 和 dp[2] ,返回之
if (i === 1 || i === 2) return i;
// 若存在記錄 dp[i] ,則直接返回之
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);
// 記錄 dp[i]
mem[i] = count;
return count;
}
/* 爬樓梯:記憶化搜尋 */
function climbingStairsDFSMem(n) {
// mem[i] 記錄爬到第 i 階的方案總數,-1 代表無記錄
const mem = new Array(n + 1).fill(-1);
return dfs(n, mem);
}
```
=== "TS"
```typescript title="climbing_stairs_dfs_mem.ts"
/* 記憶化搜尋 */
function dfs(i: number, mem: number[]): number {
// 已知 dp[1] 和 dp[2] ,返回之
if (i === 1 || i === 2) return i;
// 若存在記錄 dp[i] ,則直接返回之
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);
// 記錄 dp[i]
mem[i] = count;
return count;
}
/* 爬樓梯:記憶化搜尋 */
function climbingStairsDFSMem(n: number): number {
// mem[i] 記錄爬到第 i 階的方案總數,-1 代表無記錄
const mem = new Array(n + 1).fill(-1);
return dfs(n, mem);
}
```
=== "Dart"
```dart title="climbing_stairs_dfs_mem.dart"
/* 記憶化搜尋 */
int dfs(int i, List<int> mem) {
// 已知 dp[1] 和 dp[2] ,返回之
if (i == 1 || i == 2) return i;
// 若存在記錄 dp[i] ,則直接返回之
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);
// 記錄 dp[i]
mem[i] = count;
return count;
}
/* 爬樓梯:記憶化搜尋 */
int climbingStairsDFSMem(int n) {
// mem[i] 記錄爬到第 i 階的方案總數,-1 代表無記錄
List<int> mem = List.filled(n + 1, -1);
return dfs(n, mem);
}
```
=== "Rust"
```rust title="climbing_stairs_dfs_mem.rs"
/* 記憶化搜尋 */
fn dfs(i: usize, mem: &mut [i32]) -> i32 {
// 已知 dp[1] 和 dp[2] ,返回之
if i == 1 || i == 2 {
return i as i32;
}
// 若存在記錄 dp[i] ,則直接返回之
if mem[i] != -1 {
return mem[i];
}
// dp[i] = dp[i-1] + dp[i-2]
let count = dfs(i - 1, mem) + dfs(i - 2, mem);
// 記錄 dp[i]
mem[i] = count;
count
}
/* 爬樓梯:記憶化搜尋 */
fn climbing_stairs_dfs_mem(n: usize) -> i32 {
// mem[i] 記錄爬到第 i 階的方案總數,-1 代表無記錄
let mut mem = vec![-1; n + 1];
dfs(n, &mut mem)
}
```
=== "C"
```c title="climbing_stairs_dfs_mem.c"
/* 記憶化搜尋 */
int dfs(int i, int *mem) {
// 已知 dp[1] 和 dp[2] ,返回之
if (i == 1 || i == 2)
return i;
// 若存在記錄 dp[i] ,則直接返回之
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);
// 記錄 dp[i]
mem[i] = count;
return count;
}
/* 爬樓梯:記憶化搜尋 */
int climbingStairsDFSMem(int n) {
// mem[i] 記錄爬到第 i 階的方案總數,-1 代表無記錄
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;
}
```
=== "Kotlin"
```kotlin title="climbing_stairs_dfs_mem.kt"
/* 記憶化搜尋 */
fun dfs(i: Int, mem: IntArray): Int {
// 已知 dp[1] 和 dp[2] ,返回之
if (i == 1 || i == 2) return i
// 若存在記錄 dp[i] ,則直接返回之
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)
// 記錄 dp[i]
mem[i] = count
return count
}
/* 爬樓梯:記憶化搜尋 */
fun climbingStairsDFSMem(n: Int): Int {
// mem[i] 記錄爬到第 i 階的方案總數,-1 代表無記錄
val mem = IntArray(n + 1)
2024-04-11 01:11:20 +08:00
mem.fill(-1)
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return dfs(n, mem)
}
```
=== "Ruby"
```ruby title="climbing_stairs_dfs_mem.rb"
[class]{}-[func]{dfs}
[class]{}-[func]{climbing_stairs_dfs_mem}
```
=== "Zig"
```zig title="climbing_stairs_dfs_mem.zig"
// 記憶化搜尋
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);
}
// 若存在記錄 dp[i] ,則直接返回之
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);
// 記錄 dp[i]
mem[i] = count;
return count;
}
// 爬樓梯:記憶化搜尋
fn climbingStairsDFSMem(comptime n: usize) i32 {
// mem[i] 記錄爬到第 i 階的方案總數,-1 代表無記錄
var mem = [_]i32{ -1 } ** (n + 1);
return dfs(n, &mem);
}
```
??? pythontutor "視覺化執行"
2024-04-11 01:11:20 +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%2C%20mem%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E8%A8%98%E6%86%B6%E5%8C%96%E6%90%9C%E5%B0%8B%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%A8%98%E9%8C%84%20dp%5Bi%5D%20%EF%BC%8C%E5%89%87%E7%9B%B4%E6%8E%A5%E8%BF%94%E5%9B%9E%E4%B9%8B%0A%20%20%20%20if%20mem%5Bi%5D%20%21%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%2C%20mem%29%20%2B%20dfs%28i%20-%202%2C%20mem%29%0A%20%20%20%20%23%20%E8%A8%98%E9%8C%84%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%A8%93%E6%A2%AF%EF%BC%9A%E8%A8%98%E6%86%B6%E5%8C%96%E6%90%9C%E5%B0%8B%22%22%22%0A%20%20%20%20%23%20mem%5Bi%5D%20%E8%A8%98%E9%8C%84%E7%88%AC%E5%88%B0%E7%AC%AC%20i%20%E9%9A%8E%E7%9A%84%E6%96%B9%E6%A1%88%E7%B8%BD%E6%95%B8%EF%BC%8C-1%20%E4%BB%A3%E8%A1%A8%E7%84%A1%E8%A8%98%E9%8C%84%0A%20%20%20%20mem%20%3D%20%5B-1%5D%20%2A%20%28n%20%2B%201%29%0A%20%20%20%20return%20dfs%28n%2C%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%9A%8E%E6%A8%93%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A8%AE%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>
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2024-04-06 03:02:20 +08:00
觀察圖 14-4 **經過記憶化處理後,所有重疊子問題都只需計算一次,時間複雜度最佳化至 $O(n)$** ,這是一個巨大的飛躍。
![記憶化搜尋對應遞迴樹](intro_to_dynamic_programming.assets/climbing_stairs_dfs_memo_tree.png){ class="animation-figure" }
<p align="center"> 圖 14-4 &nbsp; 記憶化搜尋對應遞迴樹 </p>
## 14.1.3 &nbsp; 方法三:動態規劃
**記憶化搜尋是一種“從頂至底”的方法**:我們從原問題(根節點)開始,遞迴地將較大子問題分解為較小子問題,直至解已知的最小子問題(葉節點)。之後,透過回溯逐層收集子問題的解,構建出原問題的解。
與之相反,**動態規劃是一種“從底至頂”的方法**:從最小子問題的解開始,迭代地構建更大子問題的解,直至得到原問題的解。
由於動態規劃不包含回溯過程,因此只需使用迴圈迭代實現,無須使用遞迴。在以下程式碼中,我們初始化一個陣列 `dp` 來儲存子問題的解,它起到了與記憶化搜尋中陣列 `mem` 相同的記錄作用:
=== "Python"
```python title="climbing_stairs_dp.py"
def climbing_stairs_dp(n: int) -> int:
"""爬樓梯:動態規劃"""
if n == 1 or n == 2:
return n
# 初始化 dp 表,用於儲存子問題的解
dp = [0] * (n + 1)
# 初始狀態:預設最小子問題的解
dp[1], dp[2] = 1, 2
# 狀態轉移:從較小子問題逐步求解較大子問題
for i in range(3, n + 1):
dp[i] = dp[i - 1] + dp[i - 2]
return dp[n]
```
=== "C++"
```cpp title="climbing_stairs_dp.cpp"
/* 爬樓梯:動態規劃 */
int climbingStairsDP(int n) {
if (n == 1 || n == 2)
return n;
// 初始化 dp 表,用於儲存子問題的解
vector<int> dp(n + 1);
// 初始狀態:預設最小子問題的解
dp[1] = 1;
dp[2] = 2;
// 狀態轉移:從較小子問題逐步求解較大子問題
for (int i = 3; i <= n; i++) {
dp[i] = dp[i - 1] + dp[i - 2];
}
return dp[n];
}
```
=== "Java"
```java title="climbing_stairs_dp.java"
/* 爬樓梯:動態規劃 */
int climbingStairsDP(int n) {
if (n == 1 || n == 2)
return n;
// 初始化 dp 表,用於儲存子問題的解
int[] dp = new int[n + 1];
// 初始狀態:預設最小子問題的解
dp[1] = 1;
dp[2] = 2;
// 狀態轉移:從較小子問題逐步求解較大子問題
for (int i = 3; i <= n; i++) {
dp[i] = dp[i - 1] + dp[i - 2];
}
return dp[n];
}
```
=== "C#"
```csharp title="climbing_stairs_dp.cs"
/* 爬樓梯:動態規劃 */
int ClimbingStairsDP(int n) {
if (n == 1 || n == 2)
return n;
// 初始化 dp 表,用於儲存子問題的解
int[] dp = new int[n + 1];
// 初始狀態:預設最小子問題的解
dp[1] = 1;
dp[2] = 2;
// 狀態轉移:從較小子問題逐步求解較大子問題
for (int i = 3; i <= n; i++) {
dp[i] = dp[i - 1] + dp[i - 2];
}
return dp[n];
}
```
=== "Go"
```go title="climbing_stairs_dp.go"
/* 爬樓梯:動態規劃 */
func climbingStairsDP(n int) int {
if n == 1 || n == 2 {
return n
}
// 初始化 dp 表,用於儲存子問題的解
dp := make([]int, n+1)
// 初始狀態:預設最小子問題的解
dp[1] = 1
dp[2] = 2
// 狀態轉移:從較小子問題逐步求解較大子問題
for i := 3; i <= n; i++ {
dp[i] = dp[i-1] + dp[i-2]
}
return dp[n]
}
```
=== "Swift"
```swift title="climbing_stairs_dp.swift"
/* 爬樓梯:動態規劃 */
func climbingStairsDP(n: Int) -> Int {
if n == 1 || n == 2 {
return n
}
// 初始化 dp 表,用於儲存子問題的解
var dp = Array(repeating: 0, count: n + 1)
// 初始狀態:預設最小子問題的解
dp[1] = 1
dp[2] = 2
// 狀態轉移:從較小子問題逐步求解較大子問題
for i in 3 ... n {
dp[i] = dp[i - 1] + dp[i - 2]
}
return dp[n]
}
```
=== "JS"
```javascript title="climbing_stairs_dp.js"
/* 爬樓梯:動態規劃 */
function climbingStairsDP(n) {
if (n === 1 || n === 2) return n;
// 初始化 dp 表,用於儲存子問題的解
const dp = new Array(n + 1).fill(-1);
// 初始狀態:預設最小子問題的解
dp[1] = 1;
dp[2] = 2;
// 狀態轉移:從較小子問題逐步求解較大子問題
for (let i = 3; i <= n; i++) {
dp[i] = dp[i - 1] + dp[i - 2];
}
return dp[n];
}
```
=== "TS"
```typescript title="climbing_stairs_dp.ts"
/* 爬樓梯:動態規劃 */
function climbingStairsDP(n: number): number {
if (n === 1 || n === 2) return n;
// 初始化 dp 表,用於儲存子問題的解
const dp = new Array(n + 1).fill(-1);
// 初始狀態:預設最小子問題的解
dp[1] = 1;
dp[2] = 2;
// 狀態轉移:從較小子問題逐步求解較大子問題
for (let i = 3; i <= n; i++) {
dp[i] = dp[i - 1] + dp[i - 2];
}
return dp[n];
}
```
=== "Dart"
```dart title="climbing_stairs_dp.dart"
/* 爬樓梯:動態規劃 */
int climbingStairsDP(int n) {
if (n == 1 || n == 2) return n;
// 初始化 dp 表,用於儲存子問題的解
List<int> dp = List.filled(n + 1, 0);
// 初始狀態:預設最小子問題的解
dp[1] = 1;
dp[2] = 2;
// 狀態轉移:從較小子問題逐步求解較大子問題
for (int i = 3; i <= n; i++) {
dp[i] = dp[i - 1] + dp[i - 2];
}
return dp[n];
}
```
=== "Rust"
```rust title="climbing_stairs_dp.rs"
/* 爬樓梯:動態規劃 */
fn climbing_stairs_dp(n: usize) -> i32 {
// 已知 dp[1] 和 dp[2] ,返回之
if n == 1 || n == 2 {
return n as i32;
}
// 初始化 dp 表,用於儲存子問題的解
let mut dp = vec![-1; n + 1];
// 初始狀態:預設最小子問題的解
dp[1] = 1;
dp[2] = 2;
// 狀態轉移:從較小子問題逐步求解較大子問題
for i in 3..=n {
dp[i] = dp[i - 1] + dp[i - 2];
}
dp[n]
}
```
=== "C"
```c title="climbing_stairs_dp.c"
/* 爬樓梯:動態規劃 */
int climbingStairsDP(int n) {
if (n == 1 || n == 2)
return n;
// 初始化 dp 表,用於儲存子問題的解
int *dp = (int *)malloc((n + 1) * sizeof(int));
// 初始狀態:預設最小子問題的解
dp[1] = 1;
dp[2] = 2;
// 狀態轉移:從較小子問題逐步求解較大子問題
for (int i = 3; i <= n; i++) {
dp[i] = dp[i - 1] + dp[i - 2];
}
int result = dp[n];
free(dp);
return result;
}
```
=== "Kotlin"
```kotlin title="climbing_stairs_dp.kt"
/* 爬樓梯:動態規劃 */
fun climbingStairsDP(n: Int): Int {
if (n == 1 || n == 2) return n
// 初始化 dp 表,用於儲存子問題的解
val dp = IntArray(n + 1)
// 初始狀態:預設最小子問題的解
dp[1] = 1
dp[2] = 2
// 狀態轉移:從較小子問題逐步求解較大子問題
for (i in 3..n) {
dp[i] = dp[i - 1] + dp[i - 2]
}
return dp[n]
}
```
=== "Ruby"
```ruby title="climbing_stairs_dp.rb"
[class]{}-[func]{climbing_stairs_dp}
```
=== "Zig"
```zig title="climbing_stairs_dp.zig"
// 爬樓梯:動態規劃
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,用於儲存子問題的解
var dp = [_]i32{-1} ** (n + 1);
// 初始狀態:預設最小子問題的解
dp[1] = 1;
dp[2] = 2;
// 狀態轉移:從較小子問題逐步求解較大子問題
for (3..n + 1) |i| {
dp[i] = dp[i - 1] + dp[i - 2];
}
return dp[n];
}
```
??? pythontutor "視覺化執行"
2024-04-11 01:11:20 +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%A8%93%E6%A2%AF%EF%BC%9A%E5%8B%95%E6%85%8B%E8%A6%8F%E5%8A%83%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%E6%96%BC%E5%84%B2%E5%AD%98%E5%AD%90%E5%95%8F%E9%A1%8C%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%20%3D%20%5B0%5D%20%2A%20%28n%20%2B%201%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E7%8B%80%E6%85%8B%EF%BC%9A%E9%A0%90%E8%A8%AD%E6%9C%80%E5%B0%8F%E5%AD%90%E5%95%8F%E9%A1%8C%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%5B1%5D%2C%20dp%5B2%5D%20%3D%201%2C%202%0A%20%20%20%20%23%20%E7%8B%80%E6%85%8B%E8%BD%89%E7%A7%BB%EF%BC%9A%E5%BE%9E%E8%BC%83%E5%B0%8F%E5%AD%90%E5%95%8F%E9%A1%8C%E9%80%90%E6%AD%A5%E6%B1%82%E8%A7%A3%E8%BC%83%E5%A4%A7%E5%AD%90%E5%95%8F%E9%A1%8C%0A%20%20%20%20for%20i%20in%20range%283%2C%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%9A%8E%E6%A8%93%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A8%AE%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%A8%93%E6%A2%AF%EF%BC%9A%E5%8B%95%E6%85%8B%E8%A6%8F%E5%8A%83%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%E6%96%BC%E5%84%B2%E5%AD%98%E5%AD%90%E5%95%8F%E9%A1%8C%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%20%3D%20%5B0%5D%20%2A%20%28n%20%2B%201%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E7%8B%80%E6%85%8B%EF%BC%9A%E9%A0%90%E8%A8%AD%E6%9C%80%E5%B0%8F%E5%AD%90%E5%95%8F%E9%A1%8C%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%5B1%5D%2C%20dp%5B2%5D%20%3D%201%2C%202%0A%20%20%20%20%23%20%E7%8B%80%E6%85%8B%E8%BD%89%E7%A7%BB%EF%BC%9A%E5%BE%9E%E8%BC%83%E5%B0%8F%E5%AD%90%E5%95%8F%E9%A1%8C%E9%80%90%E6%AD%A5%E6%B1%82%E8%A7%A3%E8%BC%83%E5%A4%A7%E5%AD%90%E5%95%8F%E9%A1%8C%0A%20%20%20%20for%20i%20in%20range%283%2C%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%9A%8E%E6%A8%93%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A8%AE%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">全螢幕觀看 ></a></div>
2024-04-06 03:02:20 +08:00
圖 14-5 模擬了以上程式碼的執行過程。
![爬樓梯的動態規劃過程](intro_to_dynamic_programming.assets/climbing_stairs_dp.png){ class="animation-figure" }
<p align="center"> 圖 14-5 &nbsp; 爬樓梯的動態規劃過程 </p>
與回溯演算法一樣,動態規劃也使用“狀態”概念來表示問題求解的特定階段,每個狀態都對應一個子問題以及相應的區域性最優解。例如,爬樓梯問題的狀態定義為當前所在樓梯階數 $i$ 。
根據以上內容,我們可以總結出動態規劃的常用術語。
- 將陣列 `dp` 稱為 <u>dp 表</u>$dp[i]$ 表示狀態 $i$ 對應子問題的解。
- 將最小子問題對應的狀態(第 $1$ 階和第 $2$ 階樓梯)稱為<u>初始狀態</u>
- 將遞推公式 $dp[i] = dp[i-1] + dp[i-2]$ 稱為<u>狀態轉移方程</u>
## 14.1.4 &nbsp; 空間最佳化
細心的讀者可能發現了,**由於 $dp[i]$ 只與 $dp[i-1]$ 和 $dp[i-2]$ 有關,因此我們無須使用一個陣列 `dp` 來儲存所有子問題的解**,而只需兩個變數滾動前進即可。程式碼如下所示:
=== "Python"
```python title="climbing_stairs_dp.py"
def climbing_stairs_dp_comp(n: int) -> int:
"""爬樓梯:空間最佳化後的動態規劃"""
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
```
=== "C++"
```cpp title="climbing_stairs_dp.cpp"
/* 爬樓梯:空間最佳化後的動態規劃 */
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;
}
```
=== "Java"
```java title="climbing_stairs_dp.java"
/* 爬樓梯:空間最佳化後的動態規劃 */
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;
}
```
=== "C#"
```csharp title="climbing_stairs_dp.cs"
/* 爬樓梯:空間最佳化後的動態規劃 */
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;
}
```
=== "Go"
```go title="climbing_stairs_dp.go"
/* 爬樓梯:空間最佳化後的動態規劃 */
func climbingStairsDPComp(n int) int {
if n == 1 || n == 2 {
return n
}
a, b := 1, 2
// 狀態轉移:從較小子問題逐步求解較大子問題
for i := 3; i <= n; i++ {
a, b = b, a+b
}
return b
}
```
=== "Swift"
```swift title="climbing_stairs_dp.swift"
/* 爬樓梯:空間最佳化後的動態規劃 */
func climbingStairsDPComp(n: Int) -> Int {
if n == 1 || n == 2 {
return n
}
var a = 1
var b = 2
for _ in 3 ... n {
(a, b) = (b, a + b)
}
return b
}
```
=== "JS"
```javascript title="climbing_stairs_dp.js"
/* 爬樓梯:空間最佳化後的動態規劃 */
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;
}
```
=== "TS"
```typescript title="climbing_stairs_dp.ts"
/* 爬樓梯:空間最佳化後的動態規劃 */
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;
}
```
=== "Dart"
```dart title="climbing_stairs_dp.dart"
/* 爬樓梯:空間最佳化後的動態規劃 */
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;
}
```
=== "Rust"
```rust title="climbing_stairs_dp.rs"
/* 爬樓梯:空間最佳化後的動態規劃 */
fn climbing_stairs_dp_comp(n: usize) -> i32 {
if n == 1 || n == 2 {
return n as i32;
}
let (mut a, mut b) = (1, 2);
for _ in 3..=n {
let tmp = b;
b = a + b;
a = tmp;
}
b
}
```
=== "C"
```c title="climbing_stairs_dp.c"
/* 爬樓梯:空間最佳化後的動態規劃 */
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;
}
```
=== "Kotlin"
```kotlin title="climbing_stairs_dp.kt"
/* 爬樓梯:空間最佳化後的動態規劃 */
fun climbingStairsDPComp(n: Int): Int {
if (n == 1 || n == 2) return n
var a = 1
var b = 2
for (i in 3..n) {
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val temp = b
b += a
a = temp
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}
return b
}
```
=== "Ruby"
```ruby title="climbing_stairs_dp.rb"
[class]{}-[func]{climbing_stairs_dp_comp}
```
=== "Zig"
```zig title="climbing_stairs_dp.zig"
// 爬樓梯:空間最佳化後的動態規劃
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;
}
```
??? pythontutor "視覺化執行"
2024-04-11 01:11:20 +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%A8%93%E6%A2%AF%EF%BC%9A%E7%A9%BA%E9%96%93%E6%9C%80%E4%BD%B3%E5%8C%96%E5%BE%8C%E7%9A%84%E5%8B%95%E6%85%8B%E8%A6%8F%E5%8A%83%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%2C%20b%20%3D%201%2C%202%0A%20%20%20%20for%20_%20in%20range%283%2C%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20a%2C%20b%20%3D%20b%2C%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%9A%8E%E6%A8%93%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A8%AE%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%A8%93%E6%A2%AF%EF%BC%9A%E7%A9%BA%E9%96%93%E6%9C%80%E4%BD%B3%E5%8C%96%E5%BE%8C%E7%9A%84%E5%8B%95%E6%85%8B%E8%A6%8F%E5%8A%83%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%2C%20b%20%3D%201%2C%202%0A%20%20%20%20for%20_%20in%20range%283%2C%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20a%2C%20b%20%3D%20b%2C%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%9A%8E%E6%A8%93%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A8%AE%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">全螢幕觀看 ></a></div>
2024-04-06 03:02:20 +08:00
觀察以上程式碼,由於省去了陣列 `dp` 佔用的空間,因此空間複雜度從 $O(n)$ 降至 $O(1)$ 。
在動態規劃問題中,當前狀態往往僅與前面有限個狀態有關,這時我們可以只保留必要的狀態,透過“降維”來節省記憶體空間。**這種空間最佳化技巧被稱為“滾動變數”或“滾動陣列”**。