leetcode -- Edit Distance

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character
b) Delete a character
c) Replace a character

[解题思路]

这里用ed(s1, s2)来表示s1, s2之间的edit distance

• base case：

           ed("", "") = 0

           ed("", s) = ed(s, "") = ||s||

• 状态迁移方程：

           ed(s1 + ch1, s2 + ch2) = ed(s1, s2) if ch1 == ch2, 这时从s1+ch1, s2+ch2 到 s1, s2不需要做任何操作 

           

           if . Here we compare three options:

1. Replace ch1 with ch2, hence .
2. Insert ch2 into , hence .
3. Delete ch1 from , hence .

public int minDistance(String word1, String word2) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int m = word1.length(), n = word2.length();
        int[][] ed = new int[m + 1][n + 1];
        for(int i = 0; i < m + 1; i++){
            for(int j = 0; j < n + 1; j++){
                if(i == 0){
                    ed[i][j] = j;
                } else if(j == 0){
                    ed[i][j] = i;
                } else {
                    if(word1.charAt(i - 1) == word2.charAt(j - 1)){
                        ed[i][j] = ed[i - 1][j - 1];
                    } else {
                        ed[i][j] = 1 + Math.min(ed[i - 1][j - 1], Math.min(ed[i][j - 1], ed[i - 1][j]));
                    }
                }
            }
        }
        return ed[m][n];
    }

 ref:

http://tianrunhe.wordpress.com/2012/07/15/dynamic-programing-for-edit-distance-edit-distance/