LeetCode 72. Edit Distance

72. Edit Distance

Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.

You have the following 3 operations permitted on a word:

Insert a character
Delete a character
Replace a character
Example 1:

Input: word1 = “horse”, word2 = “ros”
Output: 3
Explanation:
horse -> rorse (replace ‘h’ with ‘r’)
rorse -> rose (remove ‘r’)
rose -> ros (remove ‘e’)
Example 2:

Input: word1 = “intention”, word2 = “execution”
Output: 5
Explanation:
intention -> inention (remove ‘t’)
inention -> enention (replace ‘i’ with ‘e’)
enention -> exention (replace ‘n’ with ‘x’)
exention -> exection (replace ‘n’ with ‘c’)
exection -> execution (insert ‘u’)

Approach

题目大意：给两个字符串，有三种操作增删改，求最少的操作次数。
解题思路：求次数的题，一般都要用动态规划，这里看代码注释就好了。

Code

class Solution {
public:
    int minDistance(string word1, string word2) {
        int n = word1.size(), m = word2.size();
        vector> op(n + 1, vector(m + 1, INT32_MAX));
        //dp[i][j] 表示从长度为i的字符串变为长度为j字符串最少的操作次数。
        for (int i = 0; i <= n; i++) {
            op[i][0] = i; 
        }
        for (int i = 0; i <= m; i++) {
            op[0][i] = i;
        }
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                if (word1[i - 1] == word2[j - 1]) {
                    op[i][j] = op[i - 1][j - 1];
                } else {
                    op[i][j] = min(op[i - 1][j], min(op[i][j - 1],op[i-1][j-1])) + 1; //抽象的代表三个操作增删改
                }
            }
        }
        return op[n][m];
    }
};