leetcode72

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:

1. Insert a character

2. Delete a character

3. 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')

 

DP。

定义：dp[i][j]对s, t是说s[0~i]和t[0~j]之间的edit distance是多少。

初始化：第一行第一列都按当前下标来。（即把“”变为当前这么长的substring要的代价嘛）

转移方程：

1.s[i] == s[j]时，dp[i][j] = dp[i - 1][j - 1]。（当两个字符相等时一起划掉看前面的答案，上面三种edit方法都不需要）

2.s[I] != s[j]时，dp[i][j] = 1 + min(dp[i - 1][j], dp[i][j - 1], dp[i-1][j-1]) （字母越多是越麻烦的。先用1步解决掉最后两个字母不相等的问题，再翻前面的答案本本从而找到当前解决方案所需要的以前代价。解决方法有三种：删s的最后一个，删t的最后一个，把s的替代成j的） 

 

我的实现：

class Solution {
    public int minDistance(String word1, String word2) {
        
        if (word1 == null || word2 == null) {
            return -1;
        }
        
        int[][] dp = new int[word1.length() + 1][word2.length() + 1];
        for (int j = 1; j < dp[0].length; j++) {
            dp[0][j] = j;
        }
        for (int i = 1; i < dp.length; i++) {
            dp[i][0] = i;
        }
        for (int i = 1; i < dp.length; i++) {
            for (int j = 1; j < dp[0].length; j++) {
                if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    dp[i][j] = 1 + Math.min(dp[i - 1][j], Math.min(dp[i][j - 1], dp[i - 1][j - 1]));
                }
            }
        }
        return dp[word1.length()][word2.length()];
    }
}