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 

           ed(s_{1}+ch1, s_{2}+ch2) = \min(1 + ed(s_{1}, s_{2}), 1 + ed(s_{1}+ch1, s_{2}), 1 + ed(s_{1}, s_{2}+ch2))

           if ch1 != ch2. Here we compare three options:

  1. Replace ch1 with ch2, hence 1 + ed(s_{1}, s_{2}).
  2. Insert ch2 into s_{2}, hence 1 + ed(s_{1}+ch1, s_{2}).
  3. Delete ch1 from s_{1}, hence 1 + ed(s_{1}, s_{2}+ch2).
 1 public int minDistance(String word1, String word2) {
 2         // Start typing your Java solution below
 3         // DO NOT write main() function
 4         int m = word1.length(), n = word2.length();
 5         int[][] ed = new int[m + 1][n + 1];
 6         for(int i = 0; i < m + 1; i++){
 7             for(int j = 0; j < n + 1; j++){
 8                 if(i == 0){
 9                     ed[i][j] = j;
10                 } else if(j == 0){
11                     ed[i][j] = i;
12                 } else {
13                     if(word1.charAt(i - 1) == word2.charAt(j - 1)){
14                         ed[i][j] = ed[i - 1][j - 1];
15                     } else {
16                         ed[i][j] = 1 + Math.min(ed[i - 1][j - 1], Math.min(ed[i][j - 1], ed[i - 1][j]));
17                     }
18                 }
19             }
20         }
21         return ed[m][n];
22     }

 ref:

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

posted @ 2013-08-21 16:03  feiling  阅读(597)  评论(0)  编辑  收藏  举报