【重点】【DP】72.编辑距离

题目

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)] # dp[i][j]表示word1[0,i-1]和word2[0,j-1]的最小编辑距离
        for i in range(n+1):
            dp[0][i] = i
        for i in range(m+1):
            dp[i][0] = i
        for i in range(1, m+1):
            for j in range(1, n+1):
                if word1[i-1: i] == word2[j-1: j]:
                    dp[i][j] = dp[i-1][j-1]
                else:
                    dp[i][j] = min([dp[i-1][j-1], dp[i-1][j], dp[i][j-1]]) + 1
        
        return dp[m][n]

Java

法1：DP

二维DP，自底向上，参见《算法小抄》中的解释，更容易理解！！！

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

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