给定两个单词 word1 和 word2，计算出将 word1 转换成 word2 所使用的最少操作数 。
你可以对一个单词进行如下三种操作：
插入一个字符 删除一个字符 替换一个字符 示例 1:
输入: word1 = “horse”, word2 = “ros” 输出: 3 解释: horse -> rorse (将 ‘h’ 替换为 ‘r’) rorse -> rose (删除 ‘r’) rose -> ros (删除 ‘e’) 示例 2:
输入: word1 = “intention”, word2 = “execution” 输出: 5 解释: intention -> inention (删除 ‘t’) inention -> enention (将 ‘i’ 替换为 ‘e’) enention -> exention (将 ‘n’ 替换为 ‘x’) exention -> exection (将 ‘n’ 替换为 ‘c’) exection -> execution (插入 ‘u’)
*/C++ 解法 class Solution { public: int minDistance(string word1, string word2) { int m = word1.length(); int n = word2.length(); vector<vector<int>> cost(m+1 ,vector<int>(n+1)); for( int i = 0; i <= m; i++){ cost[i][0] = i; } for(int j = 0; j<=n; j++){ cost[0][j] = j; } for(int i=1; i<=m; i++){ for(int j =1; j<=n; j++){ if(word1[i-1] == word2[j-1]){ cost[i][j] = cost[i-1][j-1]; } else{ cost[i][j] = 1 + min(cost[i-1][j-1], min(cost[i-1][j], cost[i][j-1])); } } } return cost[m][n]; } }; */Python 解法 class Solution: def minDistance(self, word1: str, word2: str) -> int: n1, n2 = len(word1), len(word2) dp = [[0]*(n2 + 1) for _ in range(n1+1)]; dp[0][0] = 0 for i in range(1,n1+1): dp[i][0]=i for j in range(1,n2+1): dp[0][j]=j for i in range(1,n1+1): for j in range(1,n2+1): if word1[i-1] == word2[j-1]: dp[i][j] = dp[i-1][j-1] else: dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1])+1 return dp[n1][n2]