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# 初探动态规划
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「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。
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在本节中,我们从一个经典例题入手,先给出它的暴力回溯解法,观察其中包含的重叠子问题,再逐步导出更高效的动态规划解法。
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!!! question "爬楼梯"
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给定一个共有 $n$ 阶的楼梯,你每步可以上 $1$ 阶或者 $2$ 阶,请问有多少种方案可以爬到楼顶。
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如下图所示,对于一个 $3$ 阶楼梯,共有 $3$ 种方案可以爬到楼顶。
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![爬到第 3 阶的方案数量](intro_to_dynamic_programming.assets/climbing_stairs_example.png)
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本题的目标是求解方案数量,**我们可以考虑通过回溯来穷举所有可能性**。具体来说,将爬楼梯想象为一个多轮选择的过程:从地面出发,每轮选择上 $1$ 阶或 $2$ 阶,每当到达楼梯顶部时就将方案数量加 $1$ ,当越过楼梯顶部时就将其剪枝。
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=== "Java"
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```java title="climbing_stairs_backtrack.java"
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[class]{climbing_stairs_backtrack}-[func]{backtrack}
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[class]{climbing_stairs_backtrack}-[func]{climbingStairsBacktrack}
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```
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=== "C++"
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```cpp title="climbing_stairs_backtrack.cpp"
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[class]{}-[func]{backtrack}
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[class]{}-[func]{climbingStairsBacktrack}
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```
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=== "Python"
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```python title="climbing_stairs_backtrack.py"
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[class]{}-[func]{backtrack}
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[class]{}-[func]{climbing_stairs_backtrack}
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```
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=== "Go"
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```go title="climbing_stairs_backtrack.go"
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[class]{}-[func]{backtrack}
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[class]{}-[func]{climbingStairsBacktrack}
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```
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=== "JS"
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```javascript title="climbing_stairs_backtrack.js"
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[class]{}-[func]{backtrack}
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[class]{}-[func]{climbingStairsBacktrack}
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```
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=== "TS"
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```typescript title="climbing_stairs_backtrack.ts"
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[class]{}-[func]{backtrack}
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[class]{}-[func]{climbingStairsBacktrack}
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```
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=== "C"
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```c title="climbing_stairs_backtrack.c"
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[class]{}-[func]{backtrack}
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[class]{}-[func]{climbingStairsBacktrack}
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```
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=== "C#"
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```csharp title="climbing_stairs_backtrack.cs"
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[class]{climbing_stairs_backtrack}-[func]{backtrack}
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[class]{climbing_stairs_backtrack}-[func]{climbingStairsBacktrack}
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```
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=== "Swift"
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```swift title="climbing_stairs_backtrack.swift"
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[class]{}-[func]{backtrack}
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[class]{}-[func]{climbingStairsBacktrack}
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```
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=== "Zig"
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```zig title="climbing_stairs_backtrack.zig"
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[class]{}-[func]{backtrack}
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[class]{}-[func]{climbingStairsBacktrack}
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```
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=== "Dart"
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```dart title="climbing_stairs_backtrack.dart"
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[class]{}-[func]{backtrack}
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[class]{}-[func]{climbingStairsBacktrack}
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```
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=== "Rust"
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```rust title="climbing_stairs_backtrack.rs"
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[class]{}-[func]{backtrack}
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[class]{}-[func]{climbing_stairs_backtrack}
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```
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## 方法一:暴力搜索
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回溯算法通常并不显式地对问题进行拆解,而是将问题看作一系列决策步骤,通过试探和剪枝,搜索所有可能的解。
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我们可以尝试从问题分解的角度分析这道题。设爬到第 $i$ 阶共有 $dp[i]$ 种方案,那么 $dp[i]$ 就是原问题,其子问题包括:
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$$
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dp[i-1] , dp[i-2] , \cdots , dp[2] , dp[1]
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$$
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由于每轮只能上 $1$ 阶或 $2$ 阶,因此当我们站在第 $i$ 阶楼梯上时,上一轮只可能站在第 $i - 1$ 阶或第 $i - 2$ 阶上。换句话说,我们只能从第 $i -1$ 阶或第 $i - 2$ 阶前往第 $i$ 阶。
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由此便可得出一个重要推论:**爬到第 $i - 1$ 阶的方案数加上爬到第 $i - 2$ 阶的方案数就等于爬到第 $i$ 阶的方案数**。公式如下:
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$$
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dp[i] = dp[i-1] + dp[i-2]
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$$
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这意味着在爬楼梯问题中,各个子问题之间存在递推关系,**原问题的解可以由子问题的解构建得来**。
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![方案数量递推关系](intro_to_dynamic_programming.assets/climbing_stairs_state_transfer.png)
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我们可以根据递推公式得到暴力搜索解法:
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- 以 $dp[n]$ 为起始点,**递归地将一个较大问题拆解为两个较小问题的和**,直至到达最小子问题 $dp[1]$ 和 $dp[2]$ 时返回。
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- 最小子问题的解 $dp[1] = 1$ , $dp[2] = 2$ 是已知的,代表爬到第 $1$ , $2$ 阶分别有 $1$ , $2$ 种方案。
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观察以下代码,它和标准回溯代码都属于深度优先搜索,但更加简洁。
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=== "Java"
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```java title="climbing_stairs_dfs.java"
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[class]{climbing_stairs_dfs}-[func]{dfs}
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[class]{climbing_stairs_dfs}-[func]{climbingStairsDFS}
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```
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=== "C++"
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```cpp title="climbing_stairs_dfs.cpp"
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[class]{}-[func]{dfs}
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[class]{}-[func]{climbingStairsDFS}
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```
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=== "Python"
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```python title="climbing_stairs_dfs.py"
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[class]{}-[func]{dfs}
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[class]{}-[func]{climbing_stairs_dfs}
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```
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=== "Go"
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```go title="climbing_stairs_dfs.go"
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[class]{}-[func]{dfs}
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[class]{}-[func]{climbingStairsDFS}
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```
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=== "JS"
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```javascript title="climbing_stairs_dfs.js"
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[class]{}-[func]{dfs}
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[class]{}-[func]{climbingStairsDFS}
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```
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=== "TS"
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```typescript title="climbing_stairs_dfs.ts"
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[class]{}-[func]{dfs}
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[class]{}-[func]{climbingStairsDFS}
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```
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=== "C"
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```c title="climbing_stairs_dfs.c"
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[class]{}-[func]{dfs}
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[class]{}-[func]{climbingStairsDFS}
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```
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=== "C#"
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```csharp title="climbing_stairs_dfs.cs"
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[class]{climbing_stairs_dfs}-[func]{dfs}
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[class]{climbing_stairs_dfs}-[func]{climbingStairsDFS}
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```
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=== "Swift"
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```swift title="climbing_stairs_dfs.swift"
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[class]{}-[func]{dfs}
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[class]{}-[func]{climbingStairsDFS}
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```
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=== "Zig"
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```zig title="climbing_stairs_dfs.zig"
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[class]{}-[func]{dfs}
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[class]{}-[func]{climbingStairsDFS}
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```
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=== "Dart"
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```dart title="climbing_stairs_dfs.dart"
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[class]{}-[func]{dfs}
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[class]{}-[func]{climbingStairsDFS}
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```
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=== "Rust"
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```rust title="climbing_stairs_dfs.rs"
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[class]{}-[func]{dfs}
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[class]{}-[func]{climbing_stairs_dfs}
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```
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下图展示了暴力搜索形成的递归树。对于问题 $dp[n]$ ,其递归树的深度为 $n$ ,时间复杂度为 $O(2^n)$ 。指数阶属于爆炸式增长,如果我们输入一个比较大的 $n$ ,则会陷入漫长的等待之中。
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![爬楼梯对应递归树](intro_to_dynamic_programming.assets/climbing_stairs_dfs_tree.png)
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观察上图发现,**指数阶的时间复杂度是由于「重叠子问题」导致的**。例如:$dp[9]$ 被分解为 $dp[8]$ 和 $dp[7]$ ,$dp[8]$ 被分解为 $dp[7]$ 和 $dp[6]$ ,两者都包含子问题 $dp[7]$ 。
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以此类推,子问题中包含更小的重叠子问题,子子孙孙无穷尽也。绝大部分计算资源都浪费在这些重叠的问题上。
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## 方法二:记忆化搜索
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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,并在搜索过程中这样做:
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1. 当首次计算 $dp[i]$ 时,我们将其记录至 `mem[i]` ,以便之后使用。
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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,从而将重叠子问题剪枝。
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=== "Java"
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```java title="climbing_stairs_dfs_mem.java"
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[class]{climbing_stairs_dfs_mem}-[func]{dfs}
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[class]{climbing_stairs_dfs_mem}-[func]{climbingStairsDFSMem}
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```
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=== "C++"
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```cpp title="climbing_stairs_dfs_mem.cpp"
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[class]{}-[func]{dfs}
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[class]{}-[func]{climbingStairsDFSMem}
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```
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=== "Python"
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|
```python title="climbing_stairs_dfs_mem.py"
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[class]{}-[func]{dfs}
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[class]{}-[func]{climbing_stairs_dfs_mem}
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```
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=== "Go"
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|
```go title="climbing_stairs_dfs_mem.go"
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[class]{}-[func]{dfsMem}
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[class]{}-[func]{climbingStairsDFSMem}
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```
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=== "JS"
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|
```javascript title="climbing_stairs_dfs_mem.js"
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[class]{}-[func]{dfs}
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[class]{}-[func]{climbingStairsDFSMem}
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```
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=== "TS"
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|
```typescript title="climbing_stairs_dfs_mem.ts"
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[class]{}-[func]{dfs}
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[class]{}-[func]{climbingStairsDFSMem}
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```
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=== "C"
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|
```c title="climbing_stairs_dfs_mem.c"
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[class]{}-[func]{dfs}
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[class]{}-[func]{climbingStairsDFSMem}
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```
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=== "C#"
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|
|
```csharp title="climbing_stairs_dfs_mem.cs"
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|
[class]{climbing_stairs_dfs_mem}-[func]{dfs}
|
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|
|
[class]{climbing_stairs_dfs_mem}-[func]{climbingStairsDFSMem}
|
|
|
|
|
```
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|
|
=== "Swift"
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|
|
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|
|
|
```swift title="climbing_stairs_dfs_mem.swift"
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|
|
|
[class]{}-[func]{dfs}
|
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|
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|
|
[class]{}-[func]{climbingStairsDFSMem}
|
|
|
|
|
```
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|
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|
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|
|
|
|
=== "Zig"
|
|
|
|
|
|
|
|
|
|
```zig title="climbing_stairs_dfs_mem.zig"
|
|
|
|
|
[class]{}-[func]{dfs}
|
|
|
|
|
|
|
|
|
|
[class]{}-[func]{climbingStairsDFSMem}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
|
|
|
|
```dart title="climbing_stairs_dfs_mem.dart"
|
|
|
|
|
[class]{}-[func]{dfs}
|
|
|
|
|
|
|
|
|
|
[class]{}-[func]{climbingStairsDFSMem}
|
|
|
|
|
```
|
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|
|
|
|
|
|
|
|
=== "Rust"
|
|
|
|
|
|
|
|
|
|
```rust title="climbing_stairs_dfs_mem.rs"
|
|
|
|
|
[class]{}-[func]{dfs}
|
|
|
|
|
|
|
|
|
|
[class]{}-[func]{climbing_stairs_dfs_mem}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
观察下图,**经过记忆化处理后,所有重叠子问题都只需被计算一次,时间复杂度被优化至 $O(n)$** ,这是一个巨大的飞跃。
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|
|
![记忆化搜索对应递归树](intro_to_dynamic_programming.assets/climbing_stairs_dfs_memo_tree.png)
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|
|
## 方法三:动态规划
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**记忆化搜索是一种“从顶至底”的方法**:我们从原问题(根节点)开始,递归地将较大子问题分解为较小子问题,直至解已知的最小子问题(叶节点)。之后,通过回溯将子问题的解逐层收集,构建出原问题的解。
|
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与之相反,**动态规划是一种“从底至顶”的方法**:从最小子问题的解开始,迭代地构建更大子问题的解,直至得到原问题的解。
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由于动态规划不包含回溯过程,因此只需使用循环迭代实现,无须使用递归。在以下代码中,我们初始化一个数组 `dp` 来存储子问题的解,它起到了记忆化搜索中数组 `mem` 相同的记录作用。
|
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|
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
|
|
|
|
```java title="climbing_stairs_dp.java"
|
|
|
|
|
[class]{climbing_stairs_dp}-[func]{climbingStairsDP}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
|
|
|
|
```cpp title="climbing_stairs_dp.cpp"
|
|
|
|
|
[class]{}-[func]{climbingStairsDP}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
|
|
|
|
```python title="climbing_stairs_dp.py"
|
|
|
|
|
[class]{}-[func]{climbing_stairs_dp}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
|
|
|
|
```go title="climbing_stairs_dp.go"
|
|
|
|
|
[class]{}-[func]{climbingStairsDP}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "JS"
|
|
|
|
|
|
|
|
|
|
```javascript title="climbing_stairs_dp.js"
|
|
|
|
|
[class]{}-[func]{climbingStairsDP}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "TS"
|
|
|
|
|
|
|
|
|
|
```typescript title="climbing_stairs_dp.ts"
|
|
|
|
|
[class]{}-[func]{climbingStairsDP}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
|
|
|
|
```c title="climbing_stairs_dp.c"
|
|
|
|
|
[class]{}-[func]{climbingStairsDP}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
|
|
|
|
```csharp title="climbing_stairs_dp.cs"
|
|
|
|
|
[class]{climbing_stairs_dp}-[func]{climbingStairsDP}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
|
|
|
|
```swift title="climbing_stairs_dp.swift"
|
|
|
|
|
[class]{}-[func]{climbingStairsDP}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
|
|
|
|
```zig title="climbing_stairs_dp.zig"
|
|
|
|
|
[class]{}-[func]{climbingStairsDP}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
|
|
|
|
```dart title="climbing_stairs_dp.dart"
|
|
|
|
|
[class]{}-[func]{climbingStairsDP}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Rust"
|
|
|
|
|
|
|
|
|
|
```rust title="climbing_stairs_dp.rs"
|
|
|
|
|
[class]{}-[func]{climbing_stairs_dp}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
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某个特定阶段,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 $i$ 。
|
|
|
|
|
|
|
|
|
|
总结以上,动态规划的常用术语包括:
|
|
|
|
|
|
|
|
|
|
- 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(即第 $1$ , $2$ 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」。
|
|
|
|
|
|
|
|
|
|
![爬楼梯的动态规划过程](intro_to_dynamic_programming.assets/climbing_stairs_dp.png)
|
|
|
|
|
|
|
|
|
|
## 状态压缩
|
|
|
|
|
|
|
|
|
|
细心的你可能发现,**由于 $dp[i]$ 只与 $dp[i-1]$ 和 $dp[i-2]$ 有关,因此我们无须使用一个数组 `dp` 来存储所有子问题的解**,而只需两个变量滚动前进即可。
|
|
|
|
|
|
|
|
|
|
=== "Java"
|
|
|
|
|
|
|
|
|
|
```java title="climbing_stairs_dp.java"
|
|
|
|
|
[class]{climbing_stairs_dp}-[func]{climbingStairsDPComp}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
|
|
|
|
```cpp title="climbing_stairs_dp.cpp"
|
|
|
|
|
[class]{}-[func]{climbingStairsDPComp}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
|
|
|
|
```python title="climbing_stairs_dp.py"
|
|
|
|
|
[class]{}-[func]{climbing_stairs_dp_comp}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
|
|
|
|
```go title="climbing_stairs_dp.go"
|
|
|
|
|
[class]{}-[func]{climbingStairsDPComp}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "JS"
|
|
|
|
|
|
|
|
|
|
```javascript title="climbing_stairs_dp.js"
|
|
|
|
|
[class]{}-[func]{climbingStairsDPComp}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "TS"
|
|
|
|
|
|
|
|
|
|
```typescript title="climbing_stairs_dp.ts"
|
|
|
|
|
[class]{}-[func]{climbingStairsDPComp}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
|
|
|
|
```c title="climbing_stairs_dp.c"
|
|
|
|
|
[class]{}-[func]{climbingStairsDPComp}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
|
|
|
|
```csharp title="climbing_stairs_dp.cs"
|
|
|
|
|
[class]{climbing_stairs_dp}-[func]{climbingStairsDPComp}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
|
|
|
|
```swift title="climbing_stairs_dp.swift"
|
|
|
|
|
[class]{}-[func]{climbingStairsDPComp}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
|
|
|
|
```zig title="climbing_stairs_dp.zig"
|
|
|
|
|
[class]{}-[func]{climbingStairsDPComp}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
|
|
|
|
```dart title="climbing_stairs_dp.dart"
|
|
|
|
|
[class]{}-[func]{climbingStairsDPComp}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Rust"
|
|
|
|
|
|
|
|
|
|
```rust title="climbing_stairs_dp.rs"
|
|
|
|
|
[class]{}-[func]{climbing_stairs_dp_comp}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
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 $O(n)$ 降低至 $O(1)$ 。
|
|
|
|
|
|
|
|
|
|
**这种空间优化技巧被称为「状态压缩」**。在常见的动态规划问题中,当前状态仅与前面有限个状态有关,这时我们可以应用状态压缩,只保留必要的状态,通过“降维”来节省内存空间。
|