[LeetCode] Edit Distance（很好的DP）

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

用m*n的矩阵vector中的（i，j）存储从word1中(0...i)到word2中（0...j）的最小变化次数

矩阵中，从上到下对应删除操作，从左到右对应插入操作，从左上到右下对应Replace操作。

class Solution {
public:
    int minDistance(string word1, string word2) { //1->2
        if(word1==word2)
            return 0;
        int len1 = word1.size();
        int len2 = word2.size();
        vector<int> temp(len2+1,0);
        vector<vector<int> > vec(len1+1,temp);

        for(int i=1;i<=len1;i++){
            vec[i][0] = i;
        }
        for(int j=1;j<=len2;j++){
            vec[0][j] = j;
        }
        for(int i=1;i<=len1;i++){
            for(int j=1;j<=len2;j++){
                int tem1,tem2;

                tem1 = min(vec[i-1][j]+1,vec[i][j-1]+1);
                tem2 = word1[i-1]==word2[j-1] ? vec[i-1][j-1]:vec[i-1][j-1]+1;
                vec[i][j] = min(tem1,tem2);
            }//end for
        }//end for
        return vec[len1][len2];
    }
};