---
comments: true
difficulty: 中等
edit_url: https://github.com/doocs/leetcode/edit/main/solution/0500-0599/0583.Delete%20Operation%20for%20Two%20Strings/README.md
tags:
    - 字符串
    - 动态规划
---

<!-- problem:start -->

# [583. 两个字符串的删除操作](https://leetcode.cn/problems/delete-operation-for-two-strings)

[English Version](/solution/0500-0599/0583.Delete%20Operation%20for%20Two%20Strings/README_EN.md)

## 题目描述

<!-- description:start -->

<p>给定两个单词&nbsp;<code>word1</code>&nbsp;和<meta charset="UTF-8" />&nbsp;<code>word2</code>&nbsp;，返回使得<meta charset="UTF-8" />&nbsp;<code>word1</code>&nbsp;和&nbsp;<meta charset="UTF-8" />&nbsp;<code>word2</code><em>&nbsp;</em><strong>相同</strong>所需的<strong>最小步数</strong>。</p>

<p><strong>每步&nbsp;</strong>可以删除任意一个字符串中的一个字符。</p>

<p>&nbsp;</p>

<p><strong>示例 1：</strong></p>

<pre>
<strong>输入:</strong> word1 = "sea", word2 = "eat"
<strong>输出:</strong> 2
<strong>解释:</strong> 第一步将 "sea" 变为 "ea" ，第二步将 "eat "变为 "ea"
</pre>

<p><strong>示例 &nbsp;2:</strong></p>

<pre>
<b>输入：</b>word1 = "leetcode", word2 = "etco"
<b>输出：</b>4
</pre>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>
<meta charset="UTF-8" />

<ul>
	<li><code>1 &lt;= word1.length, word2.length &lt;= 500</code></li>
	<li><code>word1</code>&nbsp;和&nbsp;<code>word2</code>&nbsp;只包含小写英文字母</li>
</ul>

<!-- description:end -->

## 解法

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### 方法一：动态规划

类似[1143. 最长公共子序列](https://github.com/doocs/leetcode/blob/main/solution/1100-1199/1143.Longest%20Common%20Subsequence/README.md)。

定义 `dp[i][j]` 表示使得 `word1[0:i-1]` 和 `word1[0:j-1]` 两个字符串相同所需执行的删除操作次数。

时间复杂度：$O(mn)$。

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#### Python3

```python
class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        m, n = len(word1), len(word2)
        dp = [[0] * (n + 1) for _ in range(m + 1)]
        for i in range(1, m + 1):
            dp[i][0] = i
        for j in range(1, n + 1):
            dp[0][j] = j
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if word1[i - 1] == word2[j - 1]:
                    dp[i][j] = dp[i - 1][j - 1]
                else:
                    dp[i][j] = 1 + min(dp[i - 1][j], dp[i][j - 1])
        return dp[-1][-1]
```

#### Java

```java
class Solution {
    public int minDistance(String word1, String word2) {
        int m = word1.length(), n = word2.length();
        int[][] dp = new int[m + 1][n + 1];
        for (int i = 1; i <= m; ++i) {
            dp[i][0] = i;
        }
        for (int j = 1; j <= n; ++j) {
            dp[0][j] = j;
        }
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    dp[i][j] = 1 + Math.min(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        return dp[m][n];
    }
}
```

#### C++

```cpp
class Solution {
public:
    int minDistance(string word1, string word2) {
        int m = word1.size(), n = word2.size();
        vector<vector<int>> dp(m + 1, vector<int>(n + 1));
        for (int i = 1; i <= m; ++i) dp[i][0] = i;
        for (int j = 1; j <= n; ++j) dp[0][j] = j;
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (word1[i - 1] == word2[j - 1])
                    dp[i][j] = dp[i - 1][j - 1];
                else
                    dp[i][j] = 1 + min(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        return dp[m][n];
    }
};
```

#### Go

```go
func minDistance(word1 string, word2 string) int {
	m, n := len(word1), len(word2)
	dp := make([][]int, m+1)
	for i := range dp {
		dp[i] = make([]int, n+1)
		dp[i][0] = i
	}
	for j := range dp[0] {
		dp[0][j] = j
	}
	for i := 1; i <= m; i++ {
		for j := 1; j <= n; j++ {
			if word1[i-1] == word2[j-1] {
				dp[i][j] = dp[i-1][j-1]
			} else {
				dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1])
			}
		}
	}
	return dp[m][n]
}
```

#### TypeScript

```ts
function minDistance(word1: string, word2: string): number {
    const m = word1.length;
    const n = word2.length;
    const dp = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
    for (let i = 1; i <= m; i++) {
        for (let j = 1; j <= n; j++) {
            if (word1[i - 1] === word2[j - 1]) {
                dp[i][j] = dp[i - 1][j - 1] + 1;
            } else {
                dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
    }
    const max = dp[m][n];
    return m - max + n - max;
}
```

#### Rust

```rust
impl Solution {
    pub fn min_distance(word1: String, word2: String) -> i32 {
        let (m, n) = (word1.len(), word2.len());
        let (word1, word2) = (word1.as_bytes(), word2.as_bytes());
        let mut dp = vec![vec![0; n + 1]; m + 1];
        for i in 1..=m {
            for j in 1..=n {
                dp[i][j] = if word1[i - 1] == word2[j - 1] {
                    dp[i - 1][j - 1] + 1
                } else {
                    dp[i - 1][j].max(dp[i][j - 1])
                };
            }
        }
        let max = dp[m][n];
        (m - max + (n - max)) as i32
    }
}
```

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