[LeetCode]Edit Distance

标签：java   leetcode   动态规划   

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

动态规划，难点在于状态的初始化

参考：LeetCode题解  戴方勤     (soulmachine@gmail.com)    https://github.com/soulmachine/leetcode

public class Solution {
    public int minDistance(String word1, String word2) {
        if(word1.length()==0||word2.length()==0) return Math.abs(word1.length()-word2.length());
    	int f[][] = new int[word1.length()][word2.length()];
    	boolean b2 = false;
    	for(int i=0;i<word2.length();i++){
    		if(word2.charAt(i)==word1.charAt(0)) b2 = true;
    		if(b2){
    			f[0][i] = i;
    		}else{
    			f[0][i] = i+1;
    		}
    	}
    	boolean b1 = false;
    	for(int i=0;i<word1.length();i++){
    		if(word1.charAt(i)==word2.charAt(0)) b1 = true;
    		if(b1){
    			f[i][0] = i;
    		}else{
    			f[i][0] = i+1;
    		}
    	}
        for(int i=1;i<word1.length();i++){
        	for(int j=1;j<word2.length();j++){
        	    //比较位置的字符相同
        		if(word1.charAt(i)==word2.charAt(j)){
        			f[i][j] = f[i-1][j-1];
        		}else{
        			int mn = Math.min(f[i-1][j-1]+1, f[i][j-1]+1);
        			f[i][j] = Math.min(mn, f[i-1][j]+1);
        		}
        	}
        }
        return f[word1.length()-1][word2.length()-1];
    }
}

[LeetCode]Edit Distance

标签：java   leetcode   动态规划   

原文地址：http://blog.csdn.net/guorudi/article/details/42779789