题目

法1：DP

本质是不带替换操作的最小编辑距离问题！！！

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];}
}