---
comments: true
difficulty: 中等
edit_url: https://github.com/doocs/leetcode/edit/main/solution/1100-1199/1143.Longest%20Common%20Subsequence/README.md
tags:
    - 字符串
    - 动态规划
---

<!-- problem:start -->

# [1143. 最长公共子序列](https://leetcode.cn/problems/longest-common-subsequence)

[English Version](/solution/1100-1199/1143.Longest%20Common%20Subsequence/README_EN.md)

## 题目描述

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<p>给定两个字符串 <code>text1</code> 和 <code>text2</code>，返回这两个字符串的最长 <strong>公共子序列</strong> 的长度。如果不存在 <strong>公共子序列</strong> ，返回 <code>0</code> 。</p>

<p>一个字符串的 <strong>子序列</strong><em> </em>是指这样一个新的字符串：它是由原字符串在不改变字符的相对顺序的情况下删除某些字符（也可以不删除任何字符）后组成的新字符串。</p>

<ul>
	<li>例如，<code>"ace"</code> 是 <code>"abcde"</code> 的子序列，但 <code>"aec"</code> 不是 <code>"abcde"</code> 的子序列。</li>
</ul>

<p>两个字符串的 <strong>公共子序列</strong> 是这两个字符串所共同拥有的子序列。</p>

<p> </p>

<p><strong>示例 1：</strong></p>

<pre>
<strong>输入：</strong>text1 = "abcde", text2 = "ace" 
<strong>输出：</strong>3  
<strong>解释：</strong>最长公共子序列是 "ace" ，它的长度为 3 。
</pre>

<p><strong>示例 2：</strong></p>

<pre>
<strong>输入：</strong>text1 = "abc", text2 = "abc"
<strong>输出：</strong>3
<strong>解释：</strong>最长公共子序列是 "abc" ，它的长度为 3 。
</pre>

<p><strong>示例 3：</strong></p>

<pre>
<strong>输入：</strong>text1 = "abc", text2 = "def"
<strong>输出：</strong>0
<strong>解释：</strong>两个字符串没有公共子序列，返回 0 。
</pre>

<p> </p>

<p><strong>提示：</strong></p>

<ul>
	<li><code>1 <= text1.length, text2.length <= 1000</code></li>
	<li><code>text1</code> 和 <code>text2</code> 仅由小写英文字符组成。</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 $text1$ 的前 $i$ 个字符和 $text2$ 的前 $j$ 个字符的最长公共子序列的长度。那么答案为 $f[m][n]$，其中 $m$ 和 $n$ 分别为 $text1$ 和 $text2$ 的长度。

如果 $text1$ 的第 $i$ 个字符和 $text2$ 的第 $j$ 个字符相同，则 $f[i][j] = f[i - 1][j - 1] + 1$；如果 $text1$ 的第 $i$ 个字符和 $text2$ 的第 $j$ 个字符不同，则 $f[i][j] = max(f[i - 1][j], f[i][j - 1])$。即状态转移方程为：

$$
f[i][j] =
\begin{cases}
f[i - 1][j - 1] + 1, & text1[i - 1] = text2[j - 1] \\
\max(f[i - 1][j], f[i][j - 1]), & text1[i - 1] \neq text2[j - 1]
\end{cases}
$$

时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别为 $text1$ 和 $text2$ 的长度。

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#### Python3

```python
class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        m, n = len(text1), len(text2)
        f = [[0] * (n + 1) for _ in range(m + 1)]
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if text1[i - 1] == text2[j - 1]:
                    f[i][j] = f[i - 1][j - 1] + 1
                else:
                    f[i][j] = max(f[i - 1][j], f[i][j - 1])
        return f[m][n]
```

#### Java

```java
class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        int m = text1.length(), n = text2.length();
        int[][] f = new int[m + 1][n + 1];
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (text1.charAt(i - 1) == text2.charAt(j - 1)) {
                    f[i][j] = f[i - 1][j - 1] + 1;
                } else {
                    f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]);
                }
            }
        }
        return f[m][n];
    }
}
```

#### C++

```cpp
class Solution {
public:
    int longestCommonSubsequence(string text1, string text2) {
        int m = text1.size(), n = text2.size();
        int f[m + 1][n + 1];
        memset(f, 0, sizeof f);
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (text1[i - 1] == text2[j - 1]) {
                    f[i][j] = f[i - 1][j - 1] + 1;
                } else {
                    f[i][j] = max(f[i - 1][j], f[i][j - 1]);
                }
            }
        }
        return f[m][n];
    }
};
```

#### Go

```go
func longestCommonSubsequence(text1 string, text2 string) int {
	m, n := len(text1), len(text2)
	f := make([][]int, m+1)
	for i := range f {
		f[i] = make([]int, n+1)
	}
	for i := 1; i <= m; i++ {
		for j := 1; j <= n; j++ {
			if text1[i-1] == text2[j-1] {
				f[i][j] = f[i-1][j-1] + 1
			} else {
				f[i][j] = max(f[i-1][j], f[i][j-1])
			}
		}
	}
	return f[m][n]
}
```

#### TypeScript

```ts
function longestCommonSubsequence(text1: string, text2: string): number {
    const m = text1.length;
    const n = text2.length;
    const f = 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 (text1[i - 1] === text2[j - 1]) {
                f[i][j] = f[i - 1][j - 1] + 1;
            } else {
                f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]);
            }
        }
    }
    return f[m][n];
}
```

#### Rust

```rust
impl Solution {
    pub fn longest_common_subsequence(text1: String, text2: String) -> i32 {
        let (m, n) = (text1.len(), text2.len());
        let (text1, text2) = (text1.as_bytes(), text2.as_bytes());
        let mut f = vec![vec![0; n + 1]; m + 1];
        for i in 1..=m {
            for j in 1..=n {
                f[i][j] = if text1[i - 1] == text2[j - 1] {
                    f[i - 1][j - 1] + 1
                } else {
                    f[i - 1][j].max(f[i][j - 1])
                };
            }
        }
        f[m][n]
    }
}
```

#### JavaScript

```js
/**
 * @param {string} text1
 * @param {string} text2
 * @return {number}
 */
var longestCommonSubsequence = function (text1, text2) {
    const m = text1.length;
    const n = text2.length;
    const f = 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 (text1[i - 1] == text2[j - 1]) {
                f[i][j] = f[i - 1][j - 1] + 1;
            } else {
                f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]);
            }
        }
    }
    return f[m][n];
};
```

#### C#

```cs
public class Solution {
    public int LongestCommonSubsequence(string text1, string text2) {
        int m = text1.Length, n = text2.Length;
        int[,] f = new int[m + 1, n + 1];
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (text1[i - 1] == text2[j - 1]) {
                    f[i, j] = f[i - 1, j - 1] + 1;
                } else {
                    f[i, j] = Math.Max(f[i - 1, j], f[i, j - 1]);
                }
            }
        }
        return f[m, n];
    }
}
```

#### Kotlin

```kotlin
class Solution {
    fun longestCommonSubsequence(text1: String, text2: String): Int {
        val m = text1.length
        val n = text2.length
        val f = Array(m + 1) { IntArray(n + 1) }
        for (i in 1..m) {
            for (j in 1..n) {
                if (text1[i - 1] == text2[j - 1]) {
                    f[i][j] = f[i - 1][j - 1] + 1
                } else {
                    f[i][j] = Math.max(f[i - 1][j], f[i][j - 1])
                }
            }
        }
        return f[m][n]
    }
}
```

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