【DP】583.两个字符串的删除操作

题目

法1：DP

本质是不带替换操作的最小编辑距离问题！！！

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(m+1):
            dp[i][0] = i
        for j in range(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] = min(dp[i][j-1], dp[i-1][j]) + 1

        return dp[m][n]

Java

class Solution {
    public int minDistance(String word1, String word2) {
        int m = word1.length() + 1, n = word2.length() + 1;
        int[][] dp = new int[m][n];
        for (int i = 1; i < n; ++i) { // 首行
            dp[0][i] = i;
        }
        for (int i = 1; i < m; ++i) { // 首列
            dp[i][0] = i;
        }
        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] = Math.min(dp[i - 1][j], dp[i][j - 1]) + 1;
                }
            }
        }

        return dp[m - 1][n - 1];
    }
}