Leetcode-Edit Distance

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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

Analysis:

We set up the state d[i][j] as the minimum number of steps to convert word1[0...i-1] to word2[0...j-1]. According to the three operations, we can convert [0..i-1] to [0..j-1] by

1. convert [0..i-2] to [0..j-1] and delete the char word1[i-1].

2. convert [0..i-1] to [0..j-2] and add the char word2[j-1].

3. convert [0..i-2] to [0..j-2] and replace the char word1[i-1] by the char word2[j-1] if they are not the same; if they are the same, then we don not need the replace operation.

Thus, we have the formula:

d[i][j] = min{ 1. d[i-1][j-1] if word1[i-1]==word2[j-1] or d[i-1][j-1]+1 if word1[i-1]!=word2[j-1]; 2. d[i-1][j]+1; 3. d[i][j-1]+1}

Solution:

 1 public class Solution {
 2     public int minDistance(String word1, String word2) {
 3         int len1 = word1.length(),len2=word2.length();
 4         int[][] d = new int[len1+1][len2+2];
 5 
 6         for (int i=0;i<=len1;i++) d[i][0]=i;
 7         for (int i=0;i<=len2;i++) d[0][i]=i;
 8 
 9         for (int i=1;i<=len1;i++)
10             for (int j=1;j<=len2;j++){
11                 if (word1.charAt(i-1)==word2.charAt(j-1))
12                     d[i][j]=d[i-1][j-1];
13                 else d[i][j]=d[i-1][j-1]+1;
14         
15                 if (d[i-1][j]+1<d[i][j]) d[i][j]=d[i-1][j]+1;
16                 if (d[i][j-1]+1<d[i][j]) d[i][j]=d[i][j-1]+1;
17 
18             }
19 
20         return d[len1][len2]; 
21     }
22 }

Leetcode-Edit Distance

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原文地址：http://www.cnblogs.com/lishiblog/p/4120304.html