给你两个单词 word1 和 word2，请你计算出将 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')
提示：
0 <= word1.length, word2.length <= 500word1和word2由小写英文字母组成
动态规划。
设 dp[i][j] 表示将 word1 前 i 个字符组成的字符串 word1[0...i-1] 转换成 word2 前 j 个字符组成的字符串 word2[0...j-1] 的最小操作次数。m, n 分别表示 word1, word2 的长度。
初始化 dp[i][0] = i（i∈[0, m]），dp[0][j] = j （j∈[0, m]）。
i, j 分别从 1 开始遍历，判断 word1[i - 1] 与 word2[j - 1] 是否相等：
- 若
word1[i - 1] == word2[j - 1]，则dp[i][j] = dp[i - 1][j - 1]。 - 若
word1[i - 1] != word2[j - 1]，则dp[i][j] = min(dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1]) + 1。其中dp[i - 1][j] + 1对应插入操作，dp[i][j - 1] + 1对应删除操作，dp[i - 1][j - 1] + 1对应替换操作。取三者的最小值即可。
递推公式如下：
最后返回 dp[m][n] 即可。
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], dp[i - 1][j - 1]) + 1 return dp[-1][-1]
class Solution { public int minDistance(String word1, String word2) { int m = word1.length(), n = word2.length(); int[][] dp = new int[m + 1][n + 1]; for (int i = 0; i <= m; ++i) { dp[i][0] = i; } for (int j = 0; j <= n; ++j) { dp[0][j] = j; } 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(Math.min(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1; } } } return dp[m][n]; } }
class Solution { public: int minDistance(string word1, string word2) { int m = word1.size(), n = word2.size(); vector<vector<int>> dp(m + 1, vector<int>(n + 1)); for (int i = 0; i <= m; ++i) { dp[i][0] = i; } for (int j = 0; j <= n; ++j) { dp[0][j] = j; } for (int i = 1; i <= m; ++i) { for (int j = 1; j <= n; ++j) { if (word1[i - 1] == word2[j - 1]) { dp[i][j] = dp[i - 1][j - 1]; } else { dp[i][j] = min(min(dp[i - 1][j], dp[i][j - 1]), dp[i - 1][j - 1]) + 1; } } } return dp[m][n]; } };
func minDistance(word1 string, word2 string) int { m, n := len(word1), len(word2) dp := make([][]int, m+1) for i := 0; i <= m; i++ { dp[i] = make([]int, n+1) dp[i][0] = i } for j := 0; j <= n; j++ { dp[0][j] = j } for i := 1; i <= m; i++ { for j := 1; j <= n; j++ { if word1[i-1] == word2[j-1] { dp[i][j] = dp[i-1][j-1] } else { dp[i][j] = min(min(dp[i-1][j], dp[i][j-1]), dp[i-1][j-1]) + 1 } } } return dp[m][n] } func min(a, b int) int { if a < b { return a } return b }