---
comments: true
difficulty: 中等
edit_url: https://github.com/doocs/leetcode/edit/main/solution/0700-0799/0712.Minimum%20ASCII%20Delete%20Sum%20for%20Two%20Strings/README.md
tags:
    - 字符串
    - 动态规划
---

<!-- problem:start -->

# [712. 两个字符串的最小ASCII删除和](https://leetcode.cn/problems/minimum-ascii-delete-sum-for-two-strings)

[English Version](/solution/0700-0799/0712.Minimum%20ASCII%20Delete%20Sum%20for%20Two%20Strings/README_EN.md)

## 题目描述

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<p>给定两个字符串<code>s1</code>&nbsp;和&nbsp;<code>s2</code>，返回 <em>使两个字符串相等所需删除字符的&nbsp;<strong>ASCII&nbsp;</strong>值的最小和&nbsp;</em>。</p>

<p>&nbsp;</p>

<p><strong>示例 1:</strong></p>

<pre>
<strong>输入:</strong> s1 = "sea", s2 = "eat"
<strong>输出:</strong> 231
<strong>解释:</strong> 在 "sea" 中删除 "s" 并将 "s" 的值(115)加入总和。
在 "eat" 中删除 "t" 并将 116 加入总和。
结束时，两个字符串相等，115 + 116 = 231 就是符合条件的最小和。
</pre>

<p><strong>示例&nbsp;2:</strong></p>

<pre>
<strong>输入:</strong> s1 = "delete", s2 = "leet"
<strong>输出:</strong> 403
<strong>解释:</strong> 在 "delete" 中删除 "dee" 字符串变成 "let"，
将 100[d]+101[e]+101[e] 加入总和。在 "leet" 中删除 "e" 将 101[e] 加入总和。
结束时，两个字符串都等于 "let"，结果即为 100+101+101+101 = 403 。
如果改为将两个字符串转换为 "lee" 或 "eet"，我们会得到 433 或 417 的结果，比答案更大。
</pre>

<p>&nbsp;</p>

<p><strong>提示:</strong></p>

<ul>
	<li><code>0 &lt;= s1.length, s2.length &lt;= 1000</code></li>
	<li><code>s1</code>&nbsp;和&nbsp;<code>s2</code>&nbsp;由小写英文字母组成</li>
</ul>

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## 解法

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### 方法一：动态规划

我们定义 $f[i][j]$ 表示使得 $s_1$ 的前 $i$ 个字符和 $s_2$ 的前 $j$ 个字符相等所需删除字符的 ASCII 值的最小和。那么答案就是 $f[m][n]$。

如果 $s_1[i-1] = s_2[j-1]$，那么 $f[i][j] = f[i-1][j-1]$。否则，我们可以删除 $s_1[i-1]$ 或者 $s_2[j-1]$ 中的一个，使得 $f[i][j]$ 达到最小。因此，状态转移方程如下：

$$
f[i][j]=
\begin{cases}
f[i-1][j-1], & s_1[i-1] = s_2[j-1] \\
min(f[i-1][j] + s_1[i-1], f[i][j-1] + s_2[j-1]), & s_1[i-1] \neq s_2[j-1]
\end{cases}
$$

初始状态为 $f[0][j] = f[0][j-1] + s_2[j-1]$, $f[i][0] = f[i-1][0] + s_1[i-1]$。

最后返回 $f[m][n]$ 即可。

时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别是 $s_1$ 和 $s_2$ 的长度。

相似题目：

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

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#### Python3

```python
class Solution:
    def minimumDeleteSum(self, s1: str, s2: str) -> int:
        m, n = len(s1), len(s2)
        f = [[0] * (n + 1) for _ in range(m + 1)]
        for i in range(1, m + 1):
            f[i][0] = f[i - 1][0] + ord(s1[i - 1])
        for j in range(1, n + 1):
            f[0][j] = f[0][j - 1] + ord(s2[j - 1])
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if s1[i - 1] == s2[j - 1]:
                    f[i][j] = f[i - 1][j - 1]
                else:
                    f[i][j] = min(
                        f[i - 1][j] + ord(s1[i - 1]), f[i][j - 1] + ord(s2[j - 1])
                    )
        return f[m][n]
```

#### Java

```java
class Solution {
    public int minimumDeleteSum(String s1, String s2) {
        int m = s1.length(), n = s2.length();
        int[][] f = new int[m + 1][n + 1];
        for (int i = 1; i <= m; ++i) {
            f[i][0] = f[i - 1][0] + s1.charAt(i - 1);
        }
        for (int j = 1; j <= n; ++j) {
            f[0][j] = f[0][j - 1] + s2.charAt(j - 1);
        }
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (s1.charAt(i - 1) == s2.charAt(j - 1)) {
                    f[i][j] = f[i - 1][j - 1];
                } else {
                    f[i][j]
                        = Math.min(f[i - 1][j] + s1.charAt(i - 1), f[i][j - 1] + s2.charAt(j - 1));
                }
            }
        }
        return f[m][n];
    }
}
```

#### C++

```cpp
class Solution {
public:
    int minimumDeleteSum(string s1, string s2) {
        int m = s1.size(), n = s2.size();
        int f[m + 1][n + 1];
        memset(f, 0, sizeof f);
        for (int i = 1; i <= m; ++i) {
            f[i][0] = f[i - 1][0] + s1[i - 1];
        }
        for (int j = 1; j <= n; ++j) {
            f[0][j] = f[0][j - 1] + s2[j - 1];
        }
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (s1[i - 1] == s2[j - 1]) {
                    f[i][j] = f[i - 1][j - 1];
                } else {
                    f[i][j] = min(f[i - 1][j] + s1[i - 1], f[i][j - 1] + s2[j - 1]);
                }
            }
        }
        return f[m][n];
    }
};
```

#### Go

```go
func minimumDeleteSum(s1 string, s2 string) int {
	m, n := len(s1), len(s2)
	f := make([][]int, m+1)
	for i := range f {
		f[i] = make([]int, n+1)
	}
	for i, c := range s1 {
		f[i+1][0] = f[i][0] + int(c)
	}
	for j, c := range s2 {
		f[0][j+1] = f[0][j] + int(c)
	}
	for i := 1; i <= m; i++ {
		for j := 1; j <= n; j++ {
			if s1[i-1] == s2[j-1] {
				f[i][j] = f[i-1][j-1]
			} else {
				f[i][j] = min(f[i-1][j]+int(s1[i-1]), f[i][j-1]+int(s2[j-1]))
			}
		}
	}
	return f[m][n]
}
```

#### TypeScript

```ts
function minimumDeleteSum(s1: string, s2: string): number {
    const m = s1.length;
    const n = s2.length;
    const f = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
    for (let i = 1; i <= m; ++i) {
        f[i][0] = f[i - 1][0] + s1[i - 1].charCodeAt(0);
    }
    for (let j = 1; j <= n; ++j) {
        f[0][j] = f[0][j - 1] + s2[j - 1].charCodeAt(0);
    }
    for (let i = 1; i <= m; ++i) {
        for (let j = 1; j <= n; ++j) {
            if (s1[i - 1] === s2[j - 1]) {
                f[i][j] = f[i - 1][j - 1];
            } else {
                f[i][j] = Math.min(
                    f[i - 1][j] + s1[i - 1].charCodeAt(0),
                    f[i][j - 1] + s2[j - 1].charCodeAt(0),
                );
            }
        }
    }
    return f[m][n];
}
```

#### JavaScript

```js
/**
 * @param {string} s1
 * @param {string} s2
 * @return {number}
 */
var minimumDeleteSum = function (s1, s2) {
    const m = s1.length;
    const n = s2.length;
    const f = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
    for (let i = 1; i <= m; ++i) {
        f[i][0] = f[i - 1][0] + s1[i - 1].charCodeAt(0);
    }
    for (let j = 1; j <= n; ++j) {
        f[0][j] = f[0][j - 1] + s2[j - 1].charCodeAt(0);
    }
    for (let i = 1; i <= m; ++i) {
        for (let j = 1; j <= n; ++j) {
            if (s1[i - 1] === s2[j - 1]) {
                f[i][j] = f[i - 1][j - 1];
            } else {
                f[i][j] = Math.min(
                    f[i - 1][j] + s1[i - 1].charCodeAt(0),
                    f[i][j - 1] + s2[j - 1].charCodeAt(0),
                );
            }
        }
    }
    return f[m][n];
};
```

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<!-- solution:end -->

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