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14.1   初探動態規劃

動態規劃dynamic programming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{ class="animation-figure" }

圖 14-1   爬到第 3 階的方案數量

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
        }
        // 嘗試:做出選擇,更新狀態
        backtrack(choices: choices, state: state + choice, n: n, res: &res)
        // 回退
    }
}

/* 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(
    choices: MutableList<Int>,
    state: Int,
    n: Int,
    res: MutableList<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 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
    val res = mutableListOf<Int>()
    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"
### 回溯 ###
def backtrack(choices, state, n, res)
  # 當爬到第 n 階時,方案數量加 1
  res[0] += 1 if state == n
  # 走訪所有選擇
  for choice in choices
    # 剪枝:不允許越過第 n 階
    next if state + choice > n

    # 嘗試:做出選擇,更新狀態
    backtrack(choices, state + choice, n, res)
  end
  # 回退
end

### 爬樓梯:回溯 ###
def climbing_stairs_backtrack(n)
  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)
  res.first
end
```

=== "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"

<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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14.1.1   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{ class="animation-figure" }

圖 14-2   方案數量遞推關係

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]dp[2] 時返回。其中,最小子問題的解是已知的,即 dp[1] = 1dp[2] = 2 ,表示爬到第 12 階分別有 12 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"
### 搜尋 ###
def dfs(i)
  # 已知 dp[1] 和 dp[2] ,返回之
  return i if i == 1 || i == 2
  # dp[i] = dp[i-1] + dp[i-2]
  dfs(i - 1) + dfs(i - 2)
end

### 爬樓梯:搜尋 ###
def climbing_stairs_dfs(n)
  dfs(n)
end
```

=== "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);
}
```

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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{ class="animation-figure" }

圖 14-3   爬樓梯對應遞迴樹

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]dp[7] dp[8] 被分解為 dp[7]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   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)
    mem.fill(-1)
    return dfs(n, mem)
}
```

=== "Ruby"

```ruby title="climbing_stairs_dfs_mem.rb"
### 記憶化搜尋 ###
def dfs(i, mem)
  # 已知 dp[1] 和 dp[2] ,返回之
  return i if i == 1 || i == 2
  # 若存在記錄 dp[i] ,則直接返回之
  return mem[i] if mem[i] != -1

  # dp[i] = dp[i-1] + dp[i-2]
  count = dfs(i - 1, mem) + dfs(i - 2, mem)
  # 記錄 dp[i]
  mem[i] = count
end

### 爬樓梯:記憶化搜尋 ###
def climbing_stairs_dfs_mem(n)
  # mem[i] 記錄爬到第 i 階的方案總數,-1 代表無記錄
  mem = Array.new(n + 1, -1)
  dfs(n, mem)
end
```

=== "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"

<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>
<div style="margin-top: 5px;"><a href="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=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">全螢幕觀看 ></a></div>

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{ class="animation-figure" }

圖 14-4   記憶化搜尋對應遞迴樹

14.1.3   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"
### 爬樓梯:動態規劃 ###
def climbing_stairs_dp(n)
  return n  if n == 1 || n == 2

  # 初始化 dp 表,用於儲存子問題的解
  dp = Array.new(n + 1, 0)
  # 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
  (3...(n + 1)).each { |i| dp[i] = dp[i - 1] + dp[i - 2] }

  dp[n]
end
```

=== "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"

<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>

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{ class="animation-figure" }

圖 14-5   爬樓梯的動態規劃過程

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 dp 表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)稱為初始狀態
  • 將遞推公式 dp[i] = dp[i-1] + dp[i-2] 稱為狀態轉移方程

14.1.4   空間最佳化

細心的讀者可能發現了,由於 dp[i] 只與 dp[i-1]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) {
        val temp = b
        b += a
        a = temp
    }
    return b
}
```

=== "Ruby"

```ruby title="climbing_stairs_dp.rb"
### 爬樓梯:空間最佳化後的動態規劃 ###
def climbing_stairs_dp_comp(n)
  return n if n == 1 || n == 2

  a, b = 1, 2
  (3...(n + 1)).each { a, b = b, a + b }

  b
end
```

=== "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"

<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>

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”