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583. 两个字符串的删除操作

题目描述

给定两个单词 word1 和 word2 ，返回使得 word1 和  word2 相同所需的最小步数。

每步 可以删除任意一个字符串中的一个字符。

 

示例 1：

输入: word1 = "sea", word2 = "eat"
输出: 2
解释: 第一步将 "sea" 变为 "ea" ，第二步将 "eat "变为 "ea"

示例  2:

输入：word1 = "leetcode", word2 = "etco"
输出：4

 

提示：

  • 1 <= word1.length, word2.length <= 500
  • word1 和 word2 只包含小写英文字母

解法

方法一：动态规划

我们定义 \(f[i][j]\) 表示使得字符串 \(\textit{word1}\) 的前 \(i\) 个字符和字符串 \(\textit{word2}\) 的前 \(j\) 个字符相同的最小删除步数。那么答案为 \(f[m][n]\)，其中 \(m\) 和 \(n\) 分别是字符串 \(\textit{word1}\) 和 \(\textit{word2}\) 的长度。

初始时，如果 \(j = 0\)，那么 \(f[i][0] = i\)；如果 \(i = 0\)，那么 \(f[0][j] = j\)。

当 \(i, j > 0\) 时，如果 \(\textit{word1}[i - 1] = \textit{word2}[j - 1]\)，那么 \(f[i][j] = f[i - 1][j - 1]\)；否则 \(f[i][j] = \min(f[i - 1][j], f[i][j - 1]) + 1\)。

最终返回 \(f[m][n]\) 即可。

时间复杂度 \(O(m \times n)\)，空间复杂度 \(O(m \times n)\)。其中 \(m\) 和 \(n\) 分别是字符串 \(\textit{word1}\) 和 \(\textit{word2}\) 的长度。

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

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

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class Solution {
public:
    int minDistance(string word1, string word2) {
        int m = word1.length(), n = word2.length();
        vector<vector<int>> f(m + 1, vector<int>(n + 1));
        for (int i = 0; i <= m; ++i) {
            f[i][0] = i;
        }
        for (int j = 0; j <= n; ++j) {
            f[0][j] = j;
        }
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                char a = word1[i - 1];
                char b = word2[j - 1];
                if (a == b) {
                    f[i][j] = f[i - 1][j - 1];
                } else {
                    f[i][j] = min(f[i - 1][j], f[i][j - 1]) + 1;
                }
            }
        }
        return f[m][n];
    }
};

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

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

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impl Solution {
    pub fn min_distance(word1: String, word2: String) -> i32 {
        let m = word1.len();
        let n = word2.len();
        let s: Vec<char> = word1.chars().collect();
        let t: Vec<char> = word2.chars().collect();
        let mut f = vec![vec![0; n + 1]; m + 1];

        for i in 0..=m {
            f[i][0] = i as i32;
        }
        for j in 0..=n {
            f[0][j] = j as i32;
        }

        for i in 1..=m {
            for j in 1..=n {
                let a = s[i - 1];
                let b = t[j - 1];
                if a == b {
                    f[i][j] = f[i - 1][j - 1];
                } else {
                    f[i][j] = std::cmp::min(f[i - 1][j], f[i][j - 1]) + 1;
                }
            }
        }
        f[m][n]
    }
}

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