首页 > 其他 > 详细

编程之美leetcode之编辑距离

时间:2014-08-18 20:35:03      阅读:350      评论:0      收藏:0      [点我收藏+]

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

思路:典型的动态规划,dp[i][j]表示word1和word2的编辑距离,当word1[i] == word2[j]时,dp[i][j] == dp[i-1][j-1],当word1[i] != word2[j]时,

dp[i][j] = min(dp[i-1][j-1],min(dp[i-1][j],dp[i][j-1]))+1

class Solution {
public:
    int minDistance(string word1, string word2) {
    	int length1 = word1.size(),length2 = word2.size(),i,j;
    	vector<vector<int> > dp(length1+1);
    	for(i = 0;i <= length1;++i)
    	{
    		vector<int> tmp(length2+1,0);
    		dp[i] = tmp;
    	}
    	for(i = 1; i <= length1;++i)dp[i][0] = i;
    	for(j = 1; j <= length2;++j)dp[0][j] = j;
    	for(i = 1; i <= length1;++i)
    	{
    		for(j = 1;j <= length2;++j)
    		{
    			if(word1[i-1] == word2[j-1])dp[i][j] = dp[i-1][j-1];
    			else dp[i][j] = min(dp[i-1][j-1],min(dp[i-1][j],dp[i][j-1]))+1;
    		}
    	}
    	return dp[length1][length2];
    }
};


编程之美leetcode之编辑距离,布布扣,bubuko.com

编程之美leetcode之编辑距离

原文:http://blog.csdn.net/fangjian1204/article/details/38664659

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!