编程之美leetcode之编辑距离

标签：edit distance   编辑距离   动态规划   编程之美   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

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

标签：edit distance   编辑距离   动态规划   编程之美   leetcode   

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