72. Edit Distance

难度: Medium

题目描述

Given two strings word1 and word2, return the minimum number of operations required to convert word1 to word2.

You have the following three 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')

 

Constraints:

• 0 <= word1.length, word2.length <= 500
• word1 and word2 consist of lowercase English letters.

解题思路

代码实现

解决方案

java

class Solution {
    public int minDistance(String word1, String word2) {

        int n = word1.length();
        int m = word2.length();
        if (n * m == 0) {
            return n + m;
        }
        int[][] dp = new int[n + 1][m + 1];

        for (int i = 0; i < n + 1; i++) {
            dp[i][0] = i;
        }
        for (int j = 0; j < m + 1; j++) {
            dp[0][j] = j;
        }

        for (int i = 1; i < n + 1; i++) {
            for (int j = 1; j < m + 1; j++) {
                int left = dp[i - 1][j] + 1;
                int down = dp[i][j - 1] + 1;
                int left_down = dp[i - 1][j - 1];
                if (word1.charAt(i - 1) != word2.charAt(j - 1)) {
                    left_down += 1;
                }
                dp[i][j] = Math.min(left, Math.min(down, left_down));
            }
        }

        return dp[n][m];

    }
}