Leetcode 072 编辑距离

题目：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’)
提示：
0 <= word1.length, word2.length <= 500
word1 和 word2 由小写英文字母组成

动态规划-C++代码

class Solution {
public:
    int minDistance(string word1, string word2) {
        int n1 = word1.length();
        int n2 = word2.length();
        vector<vector<int>> dp(n1 + 1, vector<int>(n2 + 1));
        for (int i = 1; i <= n1; ++i) {
            dp[i][0] = dp[i - 1][0] + 1;
        }
        for (int i = 1; i <= n2; ++i) {
            dp[0][i] = dp[0][i - 1] + 1;
        }
        for (int i = 1; i <= n1; ++i) {
            for (int j = 1;  j <= n2; ++j) {
                if (word1[i - 1] == word2[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    dp[i][j] = min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1;
                }
            }
        }
        return dp[n1][n2];
    }
};