Edit Distance leetcode java

题目：

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

 

题解：

处理这道题也是用动态规划。

动态数组dp[word1.length+1][word2.length+1]

dp[i][j]表示从word1前i个字符转换到word2前j个字符最少的步骤数。

假设word1现在遍历到字符x，word2遍历到字符y（word1当前遍历到的长度为i，word2为j）。

以下两种可能性：

1. x==y，那么不用做任何编辑操作，所以dp[i][j] = dp[i-1][j-1]

2. x != y

   (1) 在word1插入y， 那么dp[i][j] = dp[i][j-1] + 1

   (2) 在word1删除x， 那么dp[i][j] = dp[i-1][j] + 1

   (3) 把word1中的x用y来替换，那么dp[i][j] = dp[i-1][j-1] + 1

 最少的步骤就是取这三个中的最小值。

最后返回 dp[word1.length+1][word2.length+1] 即可。

 

代码如下：

 1 public static int minDistance(String word1, String word2) {
 2     int len1 = word1.length();
 3     int len2 = word2.length();
 4  
 5     // len1+1, len2+1, because finally return dp[len1][len2]
 6     int[][] dp = new int[len1 + 1][len2 + 1];
 7  
 8     for (int i = 0; i <= len1; i++) 
 9         dp[i][0] = i;
10     
11     for (int j = 0; j <= len2; j++) 
12         dp[0][j] = j;
13     
14  
15     //iterate though, and check last char
16     for (int i = 1; i <= len1; i++) {
17         char c1 = word1.charAt(i-1);
18         for (int j = 1; j <= len2; j++) {
19             char c2 = word2.charAt(j-1);
20  
21             //if last two chars equal
22             if (c1 == c2) {
23                 //update dp value for +1 length
24                 dp[i][j] = dp[i-1][j-1];
25             } else {
26                 int replace = dp[i-1][j-1] + 1;
27                 int insert = dp[i-1][j] + 1;
28                 int delete = dp[i][j-1] + 1;
29  
30                 int min = Math.min(replace, insert);
31                 min = Math.min(min,delete);
32                 dp[i][j] = min;
33             }
34         }
35     }
36  
37     return dp[len1][len2];
38 }

 Reference：

 http://www.programcreek.com/2013/12/edit-distance-in-java/

 http://blog.csdn.net/linhuanmars/article/details/24213795