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| 1 | +/* |
| 2 | + * @lc app=leetcode.cn id=72 lang=cpp |
| 3 | + * |
| 4 | + * [72] 编辑距离 |
| 5 | + * |
| 6 | + * https://leetcode.cn/problems/edit-distance/description/ |
| 7 | + * |
| 8 | + * algorithms |
| 9 | + * Hard (62.79%) |
| 10 | + * Likes: 2979 |
| 11 | + * Dislikes: 0 |
| 12 | + * Total Accepted: 369.4K |
| 13 | + * Total Submissions: 588.3K |
| 14 | + * Testcase Example: '"horse"\n"ros"' |
| 15 | + * |
| 16 | + * 给你两个单词 word1 和 word2, 请返回将 word1 转换成 word2 所使用的最少操作数 |
| 17 | + * 。 |
| 18 | + * |
| 19 | + * 你可以对一个单词进行如下三种操作: |
| 20 | + * |
| 21 | + * |
| 22 | + * 插入一个字符 |
| 23 | + * 删除一个字符 |
| 24 | + * 替换一个字符 |
| 25 | + * |
| 26 | + * |
| 27 | + * |
| 28 | + * |
| 29 | + * 示例 1: |
| 30 | + * |
| 31 | + * |
| 32 | + * 输入:word1 = "horse", word2 = "ros" |
| 33 | + * 输出:3 |
| 34 | + * 解释: |
| 35 | + * horse -> rorse (将 'h' 替换为 'r') |
| 36 | + * rorse -> rose (删除 'r') |
| 37 | + * rose -> ros (删除 'e') |
| 38 | + * |
| 39 | + * |
| 40 | + * 示例 2: |
| 41 | + * |
| 42 | + * |
| 43 | + * 输入:word1 = "intention", word2 = "execution" |
| 44 | + * 输出:5 |
| 45 | + * 解释: |
| 46 | + * intention -> inention (删除 't') |
| 47 | + * inention -> enention (将 'i' 替换为 'e') |
| 48 | + * enention -> exention (将 'n' 替换为 'x') |
| 49 | + * exention -> exection (将 'n' 替换为 'c') |
| 50 | + * exection -> execution (插入 'u') |
| 51 | + * |
| 52 | + * |
| 53 | + * |
| 54 | + * |
| 55 | + * 提示: |
| 56 | + * |
| 57 | + * |
| 58 | + * 0 <= word1.length, word2.length <= 500 |
| 59 | + * word1 和 word2 由小写英文字母组成 |
| 60 | + * |
| 61 | + * |
| 62 | + */ |
| 63 | + |
| 64 | +#include <string> |
| 65 | +#include <vector> |
| 66 | +using namespace std; |
| 67 | + |
| 68 | +// @lc code=start |
| 69 | +class Solution { |
| 70 | +public: |
| 71 | + // dp[i][j]表示s的前i个字符与t的前j个字符的编辑距离 |
| 72 | + // 当s[i]==t[j]时,dp[i][j] = dp[i-1][j-1] |
| 73 | + // 当s[i]!=t[j]时,dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1])+1 |
| 74 | + // 其中dp[i-1][j-1] 表示替换操作,dp[i-1][j] 表示删除操作,dp[i][j-1] |
| 75 | + // 表示插入操作 |
| 76 | + int minDistance(string word1, string word2) { |
| 77 | + int m = word1.size(), n = word2.size(); |
| 78 | + if (m * n == 0) { |
| 79 | + return m + n; |
| 80 | + } |
| 81 | + vector<vector<int>> dp(m + 1, vector<int>(n + 1)); |
| 82 | + for (int i = 0; i <= m; i++) { |
| 83 | + dp[i][0] = i; |
| 84 | + } |
| 85 | + for (int i = 0; i <= n; i++) { |
| 86 | + dp[0][i] = i; |
| 87 | + } |
| 88 | + for (int i = 1; i <= m; i++) { |
| 89 | + for (int j = 1; j <= n; j++) { |
| 90 | + if (word1[i - 1] == word2[j - 1]) { |
| 91 | + dp[i][j] = dp[i - 1][j - 1]; |
| 92 | + } else { |
| 93 | + dp[i][j] = min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1; |
| 94 | + } |
| 95 | + } |
| 96 | + } |
| 97 | + return dp[m][n]; |
| 98 | + } |
| 99 | +}; |
| 100 | +// @lc code=end |
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