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14.1   初探动态规划

动态规划dynamic programming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。

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"

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?

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。

爬到第 3 阶的方案数量{ class="animation-figure" }

图 14-1   爬到第 3 阶的方案数量

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:

=== "Python"

```python title="climbing_stairs_backtrack.py"
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:
            continue
        # 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"""
    choices = [1, 2]  # 可选择向上爬 1 阶或 2 阶
    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]
```

=== "C++"

```cpp title="climbing_stairs_backtrack.cpp"
/* 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)
            continue;
        // 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) {
    vector<int> choices = {1, 2}; // 可选择向上爬 1 阶或 2 阶
    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];
}
```

=== "Java"

```java title="climbing_stairs_backtrack.java"
/* 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)
            continue;
        // 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) {
    List<Integer> choices = Arrays.asList(1, 2); // 可选择向上爬 1 阶或 2 阶
    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);
}
```

=== "C#"

```csharp title="climbing_stairs_backtrack.cs"
/* 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
    foreach (int 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 */
int ClimbingStairsBacktrack(int n) {
    List<int> choices = [1, 2]; // 可选择向上爬 1 阶或 2 阶
    int state = 0; // 从第 0 阶开始爬
    List<int> res = [0]; // 使用 res[0] 记录方案数量
    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) {
    // 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 {
            continue
        }
        // 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 {
    // 可选择向上爬 1 阶或 2 阶
    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]
}
```

=== "Swift"

```swift title="climbing_stairs_backtrack.swift"
/* 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 {
            continue
        }
        // 尝试:做出选择,更新状态
        backtrack(choices: choices, state: state + choice, n: n, res: &res)
        // 回退
    }
}

/* 爬楼梯:回溯 */
func climbingStairsBacktrack(n: Int) -> Int {
    let choices = [1, 2] // 可选择向上爬 1 阶或 2 阶
    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]
}
```

=== "JS"

```javascript title="climbing_stairs_backtrack.js"
/* 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
        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 */
function climbingStairsBacktrack(n) {
    const choices = [1, 2]; // 可选择向上爬 1 阶或 2 阶
    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);
}
```

=== "TS"

```typescript title="climbing_stairs_backtrack.ts"
/* 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
        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 */
function climbingStairsBacktrack(n: number): number {
    const choices = [1, 2]; // 可选择向上爬 1 阶或 2 阶
    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);
}
```

=== "Dart"

```dart title="climbing_stairs_backtrack.dart"
/* 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
    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 */
int climbingStairsBacktrack(int n) {
  List<int> choices = [1, 2]; // 可选择向上爬 1 阶或 2 阶
  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];
}
```

=== "Rust"

```rust title="climbing_stairs_backtrack.rs"
/* 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
    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 */
fn climbing_stairs_backtrack(n: usize) -> i32 {
    let choices = vec![1, 2]; // 可选择向上爬 1 阶或 2 阶
    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]
}
```

=== "C"

```c title="climbing_stairs_backtrack.c"
/* 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)
            continue;
        // 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) {
    int choices[2] = {1, 2}; // 可选择向上爬 1 阶或 2 阶
    int state = 0;           // 从第 0 阶开始爬
    int *res = (int *)malloc(sizeof(int));
    *res = 0; // 使用 res[0] 记录方案数量
    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>
) {
    // 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 = mutableListOf<Int>()
    res.add(0) // 使用 res[0] 记录方案数量
    backtrack(choices, state, n, res)
    return res[0]
}
```

=== "Ruby"

```ruby title="climbing_stairs_backtrack.rb"
### 回溯 ###
def backtrack(choices, state, n, res)
  # 当爬到第 n 阶时,方案数量加 1
  res[0] += 1 if state == n
  # 遍历所有选择
  for choice in choices
    # 剪枝:不允许越过第 n 阶
    next if state + choice > n

    # 尝试:做出选择,更新状态
    backtrack(choices, state + choice, n, res)
  end
  # 回退
end

### 爬楼梯:回溯 ###
def climbing_stairs_backtrack(n)
  choices = [1, 2] # 可选择向上爬 1 阶或 2 阶
  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)
  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 {
    // 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) {
            continue;
        }
        // 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 {
    var choices = [_]i32{ 1, 2 }; // 可选择向上爬 1 阶或 2 阶
    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];
}
```

??? pythontutor "可视化运行"

<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,%20state%3A%20int,%20n%3A%20int,%20res%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%9B%9E%E6%BA%AF%22%22%22%0A%20%20%20%20%23%20%E5%BD%93%E7%88%AC%E5%88%B0%E7%AC%AC%20n%20%E9%98%B6%E6%97%B6%EF%BC%8C%E6%96%B9%E6%A1%88%E6%95%B0%E9%87%8F%E5%8A%A0%201%0A%20%20%20%20if%20state%20%3D%3D%20n%3A%0A%20%20%20%20%20%20%20%20res%5B0%5D%20%2B%3D%201%0A%20%20%20%20%23%20%E9%81%8D%E5%8E%86%E6%89%80%E6%9C%89%E9%80%89%E6%8B%A9%0A%20%20%20%20for%20choice%20in%20choices%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%EF%BC%9A%E4%B8%8D%E5%85%81%E8%AE%B8%E8%B6%8A%E8%BF%87%E7%AC%AC%20n%20%E9%98%B6%0A%20%20%20%20%20%20%20%20if%20state%20%2B%20choice%20%3E%20n%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20continue%0A%20%20%20%20%20%20%20%20%23%20%E5%B0%9D%E8%AF%95%EF%BC%9A%E5%81%9A%E5%87%BA%E9%80%89%E6%8B%A9%EF%BC%8C%E6%9B%B4%E6%96%B0%E7%8A%B6%E6%80%81%0A%20%20%20%20%20%20%20%20backtrack%28choices,%20state%20%2B%20choice,%20n,%20res%29%0A%20%20%20%20%20%20%20%20%23%20%E5%9B%9E%E9%80%80%0A%0A%0Adef%20climbing_stairs_backtrack%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E5%9B%9E%E6%BA%AF%22%22%22%0A%20%20%20%20choices%20%3D%20%5B1,%202%5D%20%20%23%20%E5%8F%AF%E9%80%89%E6%8B%A9%E5%90%91%E4%B8%8A%E7%88%AC%201%20%E9%98%B6%E6%88%96%202%20%E9%98%B6%0A%20%20%20%20state%20%3D%200%20%20%23%20%E4%BB%8E%E7%AC%AC%200%20%E9%98%B6%E5%BC%80%E5%A7%8B%E7%88%AC%0A%20%20%20%20res%20%3D%20%5B0%5D%20%20%23%20%E4%BD%BF%E7%94%A8%20res%5B0%5D%20%E8%AE%B0%E5%BD%95%E6%96%B9%E6%A1%88%E6%95%B0%E9%87%8F%0A%20%20%20%20backtrack%28choices,%20state,%20n,%20res%29%0A%20%20%20%20return%20res%5B0%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%204%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_backtrack%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
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14.1.1   方法一:暴力搜索

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。

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:


dp[i-1], dp[i-2], \dots, dp[2], dp[1]

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。

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。

方案数量递推关系{ class="animation-figure" }

图 14-2   方案数量递推关系

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]dp[2] 时返回。其中,最小子问题的解是已知的,即 dp[1] = 1dp[2] = 2 ,表示爬到第 12 阶分别有 12 种方案。

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:

=== "Python"

```python title="climbing_stairs_dfs.py"
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)
```

=== "C++"

```cpp title="climbing_stairs_dfs.cpp"
/* 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);
}
```

=== "Java"

```java title="climbing_stairs_dfs.java"
/* 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);
}
```

=== "C#"

```csharp title="climbing_stairs_dfs.cs"
/* 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);
}
```

=== "Go"

```go title="climbing_stairs_dfs.go"
/* 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)
}
```

=== "Swift"

```swift title="climbing_stairs_dfs.swift"
/* 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)
}
```

=== "JS"

```javascript title="climbing_stairs_dfs.js"
/* 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);
}
```

=== "TS"

```typescript title="climbing_stairs_dfs.ts"
/* 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);
}
```

=== "Dart"

```dart title="climbing_stairs_dfs.dart"
/* 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);
}
```

=== "Rust"

```rust title="climbing_stairs_dfs.rs"
/* 搜索 */
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
}

/* 爬楼梯:搜索 */
fn climbing_stairs_dfs(n: usize) -> i32 {
    dfs(n)
}
```

=== "C"

```c title="climbing_stairs_dfs.c"
/* 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);
}
```

=== "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)
}
```

=== "Ruby"

```ruby title="climbing_stairs_dfs.rb"
### 搜索 ###
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

### 爬楼梯:搜索 ###
def climbing_stairs_dfs(n)
  dfs(n)
end
```

=== "Zig"

```zig title="climbing_stairs_dfs.zig"
// 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);
}
```

??? pythontutor "可视化运行"

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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。

爬楼梯对应递归树{ class="animation-figure" }

图 14-3   爬楼梯对应递归树

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]dp[7] dp[8] 被分解为 dp[7]dp[6] ,两者都包含子问题 dp[7]

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。

14.1.2   方法二:记忆化搜索

为了提升算法效率,我们希望所有的重叠子问题都只被计算一次。为此,我们声明一个数组 mem 来记录每个子问题的解,并在搜索过程中将重叠子问题剪枝。

  1. 当首次计算 dp[i] 时,我们将其记录至 mem[i] ,以便之后使用。
  2. 当再次需要计算 dp[i] 时,我们便可直接从 mem[i] 中获取结果,从而避免重复计算该子问题。

代码如下所示:

=== "Python"

```python title="climbing_stairs_dfs_mem.py"
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)
```

=== "C++"

```cpp title="climbing_stairs_dfs_mem.cpp"
/* 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);
}
```

=== "Java"

```java title="climbing_stairs_dfs_mem.java"
/* 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);
}
```

=== "C#"

```csharp title="climbing_stairs_dfs_mem.cs"
/* 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];
    Array.Fill(mem, -1);
    return DFS(n, mem);
}
```

=== "Go"

```go title="climbing_stairs_dfs_mem.go"
/* 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)
}
```

=== "Swift"

```swift title="climbing_stairs_dfs_mem.swift"
/* 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)
}
```

=== "JS"

```javascript title="climbing_stairs_dfs_mem.js"
/* 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);
}
```

=== "TS"

```typescript title="climbing_stairs_dfs_mem.ts"
/* 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);
}
```

=== "Dart"

```dart title="climbing_stairs_dfs_mem.dart"
/* 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);
}
```

=== "Rust"

```rust title="climbing_stairs_dfs_mem.rs"
/* 记忆化搜索 */
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 - 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)
}
```

=== "C"

```c title="climbing_stairs_dfs_mem.c"
/* 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;
}
```

=== "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)
    mem.fill(-1)
    return dfs(n, mem)
}
```

=== "Ruby"

```ruby title="climbing_stairs_dfs_mem.rb"
### 记忆化搜索 ###
def dfs(i, mem)
  # 已知 dp[1] 和 dp[2] ,返回之
  return i if i == 1 || i == 2
  # 若存在记录 dp[i] ,则直接返回之
  return mem[i] if mem[i] != -1

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

### 爬楼梯:记忆化搜索 ###
def climbing_stairs_dfs_mem(n)
  # mem[i] 记录爬到第 i 阶的方案总数,-1 代表无记录
  mem = Array.new(n + 1, -1)
  dfs(n, mem)
end
```

=== "Zig"

```zig title="climbing_stairs_dfs_mem.zig"
// 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);
}
```

??? pythontutor "可视化运行"

<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20dfs%28i%3A%20int,%20mem%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E8%AE%B0%E5%BF%86%E5%8C%96%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20%E5%B7%B2%E7%9F%A5%20dp%5B1%5D%20%E5%92%8C%20dp%5B2%5D%20%EF%BC%8C%E8%BF%94%E5%9B%9E%E4%B9%8B%0A%20%20%20%20if%20i%20%3D%3D%201%20or%20i%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20i%0A%20%20%20%20%23%20%E8%8B%A5%E5%AD%98%E5%9C%A8%E8%AE%B0%E5%BD%95%20dp%5Bi%5D%20%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E8%BF%94%E5%9B%9E%E4%B9%8B%0A%20%20%20%20if%20mem%5Bi%5D%20!%3D%20-1%3A%0A%20%20%20%20%20%20%20%20return%20mem%5Bi%5D%0A%20%20%20%20%23%20dp%5Bi%5D%20%3D%20dp%5Bi-1%5D%20%2B%20dp%5Bi-2%5D%0A%20%20%20%20count%20%3D%20dfs%28i%20-%201,%20mem%29%20%2B%20dfs%28i%20-%202,%20mem%29%0A%20%20%20%20%23%20%E8%AE%B0%E5%BD%95%20dp%5Bi%5D%0A%20%20%20%20mem%5Bi%5D%20%3D%20count%0A%20%20%20%20return%20count%0A%0A%0Adef%20climbing_stairs_dfs_mem%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E8%AE%B0%E5%BF%86%E5%8C%96%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20mem%5Bi%5D%20%E8%AE%B0%E5%BD%95%E7%88%AC%E5%88%B0%E7%AC%AC%20i%20%E9%98%B6%E7%9A%84%E6%96%B9%E6%A1%88%E6%80%BB%E6%95%B0%EF%BC%8C-1%20%E4%BB%A3%E8%A1%A8%E6%97%A0%E8%AE%B0%E5%BD%95%0A%20%20%20%20mem%20%3D%20%5B-1%5D%20*%20%28n%20%2B%201%29%0A%20%20%20%20return%20dfs%28n,%20mem%29%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_dfs_mem%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20dfs%28i%3A%20int,%20mem%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E8%AE%B0%E5%BF%86%E5%8C%96%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20%E5%B7%B2%E7%9F%A5%20dp%5B1%5D%20%E5%92%8C%20dp%5B2%5D%20%EF%BC%8C%E8%BF%94%E5%9B%9E%E4%B9%8B%0A%20%20%20%20if%20i%20%3D%3D%201%20or%20i%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20i%0A%20%20%20%20%23%20%E8%8B%A5%E5%AD%98%E5%9C%A8%E8%AE%B0%E5%BD%95%20dp%5Bi%5D%20%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E8%BF%94%E5%9B%9E%E4%B9%8B%0A%20%20%20%20if%20mem%5Bi%5D%20!%3D%20-1%3A%0A%20%20%20%20%20%20%20%20return%20mem%5Bi%5D%0A%20%20%20%20%23%20dp%5Bi%5D%20%3D%20dp%5Bi-1%5D%20%2B%20dp%5Bi-2%5D%0A%20%20%20%20count%20%3D%20dfs%28i%20-%201,%20mem%29%20%2B%20dfs%28i%20-%202,%20mem%29%0A%20%20%20%20%23%20%E8%AE%B0%E5%BD%95%20dp%5Bi%5D%0A%20%20%20%20mem%5Bi%5D%20%3D%20count%0A%20%20%20%20return%20count%0A%0A%0Adef%20climbing_stairs_dfs_mem%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E8%AE%B0%E5%BF%86%E5%8C%96%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20mem%5Bi%5D%20%E8%AE%B0%E5%BD%95%E7%88%AC%E5%88%B0%E7%AC%AC%20i%20%E9%98%B6%E7%9A%84%E6%96%B9%E6%A1%88%E6%80%BB%E6%95%B0%EF%BC%8C-1%20%E4%BB%A3%E8%A1%A8%E6%97%A0%E8%AE%B0%E5%BD%95%0A%20%20%20%20mem%20%3D%20%5B-1%5D%20*%20%28n%20%2B%201%29%0A%20%20%20%20return%20dfs%28n,%20mem%29%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_dfs_mem%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">全屏观看 ></a></div>

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。

记忆化搜索对应递归树{ class="animation-figure" }

图 14-4   记忆化搜索对应递归树

14.1.3   方法三:动态规划

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。

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。

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:

=== "Python"

```python title="climbing_stairs_dp.py"
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]
```

=== "C++"

```cpp title="climbing_stairs_dp.cpp"
/* 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];
}
```

=== "Java"

```java title="climbing_stairs_dp.java"
/* 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];
}
```

=== "C#"

```csharp title="climbing_stairs_dp.cs"
/* 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];
}
```

=== "Go"

```go title="climbing_stairs_dp.go"
/* 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]
}
```

=== "Swift"

```swift title="climbing_stairs_dp.swift"
/* 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
    for i in 3 ... n {
        dp[i] = dp[i - 1] + dp[i - 2]
    }
    return dp[n]
}
```

=== "JS"

```javascript title="climbing_stairs_dp.js"
/* 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];
}
```

=== "TS"

```typescript title="climbing_stairs_dp.ts"
/* 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];
}
```

=== "Dart"

```dart title="climbing_stairs_dp.dart"
/* 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];
}
```

=== "Rust"

```rust title="climbing_stairs_dp.rs"
/* 爬楼梯:动态规划 */
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]
}
```

=== "C"

```c title="climbing_stairs_dp.c"
/* 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;
}
```

=== "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]
}
```

=== "Ruby"

```ruby title="climbing_stairs_dp.rb"
### 爬楼梯:动态规划 ###
def climbing_stairs_dp(n)
  return n  if n == 1 || n == 2

  # 初始化 dp 表,用于存储子问题的解
  dp = Array.new(n + 1, 0)
  # 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
  (3...(n + 1)).each { |i| dp[i] = dp[i - 1] + dp[i - 2] }

  dp[n]
end
```

=== "Zig"

```zig title="climbing_stairs_dp.zig"
// 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];
}
```

??? pythontutor "可视化运行"

<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%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>

图 14-5 模拟了以上代码的执行过程。

爬楼梯的动态规划过程{ class="animation-figure" }

图 14-5   爬楼梯的动态规划过程

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

根据以上内容,我们可以总结出动态规划的常用术语。

  • 将数组 dp 称为 dp 表dp[i] 表示状态 i 对应子问题的解。
  • 将最小子问题对应的状态(第 1 阶和第 2 阶楼梯)称为初始状态
  • 将递推公式 dp[i] = dp[i-1] + dp[i-2] 称为状态转移方程

14.1.4   空间优化

细心的读者可能发现了,由于 dp[i] 只与 dp[i-1]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:

=== "Python"

```python title="climbing_stairs_dp.py"
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
```

=== "C++"

```cpp title="climbing_stairs_dp.cpp"
/* 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;
}
```

=== "Java"

```java title="climbing_stairs_dp.java"
/* 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;
}
```

=== "C#"

```csharp title="climbing_stairs_dp.cs"
/* 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;
}
```

=== "Go"

```go title="climbing_stairs_dp.go"
/* 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
}
```

=== "Swift"

```swift title="climbing_stairs_dp.swift"
/* 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
    for _ in 3 ... n {
        (a, b) = (b, a + b)
    }
    return b
}
```

=== "JS"

```javascript title="climbing_stairs_dp.js"
/* 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;
}
```

=== "TS"

```typescript title="climbing_stairs_dp.ts"
/* 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;
}
```

=== "Dart"

```dart title="climbing_stairs_dp.dart"
/* 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;
}
```

=== "Rust"

```rust title="climbing_stairs_dp.rs"
/* 爬楼梯:空间优化后的动态规划 */
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;
}
```

=== "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 temp = b
        b += a
        a = temp
    }
    return b
}
```

=== "Ruby"

```ruby title="climbing_stairs_dp.rb"
### 爬楼梯:空间优化后的动态规划 ###
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;
}
```

??? pythontutor "可视化运行"

<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%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>

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)

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”