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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

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