LintCode-编辑距离

 1 public class Solution {
 2     /**
 3      * @param word1 & word2: Two string.
 4      * @return: The minimum number of steps.
 5      */
 6     public int minDistance(String word1, String word2) {
 7         // write your code here
 8         int[][] dp = new int[word1.length()+1][word2.length()+1];
 9         
10         for(int i=1;i<=word1.length();i++)
11             dp[i][0] = i;
12             
13         for(int i=1;i<=word2.length();i++)
14             dp[0][i] = i;
15             
16         for(int i=0;i<word1.length();i++){
17             for(int j=0;j<word2.length();j++){
18                 if(word1.charAt(i) == word2.charAt(j)){
19                     dp[i+1][j+1] = Math.min(dp[i][j],Math.min(dp[i][j+1],dp[i+1][j])+1);
20                 }else{
21                     dp[i+1][j+1] = Math.min(dp[i][j+1]+1,Math.min(dp[i][j],dp[i+1][j])+1);
22                 }
23             }
24         }
25         return dp[word1.length()][word2.length()];
26     }
27 }