0072.编辑距离.cpp

/*
 * @lc app=leetcode.cn id=72 lang=cpp
 *
 * [72] 编辑距离
 *
 * https://leetcode-cn.com/problems/edit-distance/description/
 *
 * algorithms
 * Hard (55.02%)
 * Likes:    449
 * Dislikes: 0
 * Total Accepted:    22.8K
 * Total Submissions: 41.2K
 * Testcase Example:  '"horse"\n"ros"'
 *
 * 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')
 * 
 * 
 */

// @lc code=start
#include <string>
#include <vector>
#include <algorithm>
using namespace std;

class Solution {
public:
    int minDistance(string word1, string word2) {
        int len1 = word1.size();
        int len2 = word2.size();
        vector<vector<int>> dp(len1 + 1, vector<int>(len2 + 1, 0));
        for (int i = 0; i < len1 + 1; i++) {
            dp[i][0] = i;
        }
        for (int i = 0; i < len2 + 1; i++) {
            dp[0][i] = i;
        }
        for (int i = 1; i < len1 + 1; i++) {
            for (int j = 1; j < len2 + 1; j++) {
                if (word1[i - 1] == word2[j -1]) {
                    dp[i][j] = 1 + min(min(dp[i - 1][j], dp[i][j - 1]), dp[i - 1][j - 1] - 1);
                }
                else {
                    dp[i][j] = 1 + min(min(dp[i - 1][j], dp[i][j - 1]), dp[i - 1][j - 1]);
                }
            }
        }
        return dp[len1][len2];
    }
};
// @lc code=end