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:

• Insert a character
• Delete a character
• Replace a character

Example

Given word1 = "mart" and word2 = "karma", return 3.

教科书上的DP问题 不废话 直接上code

public class Solution {
    /**
     * @param word1 & word2: Two string.
     * @return: The minimum number of steps.
     */
    public int minDistance(String word1, String word2) {
        // write your code here
        if(word1==null&&word2==null) {
            return 0;
        }
        else if(word1!=null&&(word2==null||word2.length()==0)){
            return word1.length();
        } else if ((word1==null||word1.length()==0)&&word2!=null){
            return word2.length();  
        }
        
        int len1 = word1.length();
        int len2 = word2.length();
        
        int[][] t = new int[len1][len2];
        
        int i=0;
        for(; i< len2;i++){
            if(word1.charAt(0)==word2.charAt(i)){
                t[0][i] = i;
                break;
            }else{
                t[0][i] = i+1;
            }
        }
        for(;i<len2;i++){
            t[0][i] = i;
        }
        i=0;
        for(; i< len1;i++){
            if(word2.charAt(0)==word1.charAt(i)){
                t[i][0] = i;
                break;
            }else{
                t[i][0] = i+1;
            }
        }
        for(;i<len1;i++){
            t[i][0] = i;
        }
        
        for(i=1; i< len1;i++){
            for( int j=1;j<len2;j++){
                if(word1.charAt(i)==word2.charAt(j)){
                    t[i][j] = Math.min(t[i-1][j-1], Math.min(t[i-1][j]+1, t[i][j-1]+1));
                }else{
                    t[i][j] = Math.min(t[i-1][j-1], Math.min(t[i-1][j], t[i][j-1]))+1;
                }
            }
        }
        return t[len1-1][len2-1];
    }
}