Edit Distance

关于在二维矩阵中，具体 [i-1][j]和[i][j-1]谁是delete 谁是insert，要根据你已横向字符串以基准 还是以纵向字符串为基准

因为inde从1开始，在定位string的char的时候 应该减去1，不然到最后 string会越界

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')

经典dp题

dp[i][j] 表示从第一个字符串i的位置，转换到第二个字符串j位置，需要几步

时间和空间复杂度 o(n*m)

[LeetCode] 72. Edit Distance 编辑距离 - Grandyang - 博客园www.cnblogs.com

https://www.youtube.com/watch?v=iHtd2KjWX_cwww.youtube.com

class Solution {
    public int minDistance(String word1, String word2) {
    
        if(word1.length() == 0 ){
            return word2.length();
        }
        else if(word2.length() == 0){
            return word1.length();
        }
        int n1 = word1.length(), n2 = word2.length();
        
        int[][] dp = new int[n1+1][n2+1];
        
        for(int i = 0; i <= n2;i++){
            dp[0][i] = i;
        }
        
        for(int i = 0; i <= n1; i++){
            dp[i][0] = i;
        }

        for(int i = 1; i <=n1;i++){
            for(int j = 1; j <=n2 ; j++){
            
            //注意index
                if(word1.charAt(i-1) == word2.charAt(j-1)){
                    dp[i][j] = dp[i-1][j-1];
                }
                
                else{
                    dp[i][j] = Math.min(Math.min(dp[i][j-1],dp[i-1][j]),dp[i-1][j-1])+1;
                }
            }
        }
        
        return dp[n1][n2];
    }
}