[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

 

这题真是火大啊，犯了个低级错误，在private里开了个f[1000][1000]，然后每次都memset(f, 0, sizeof(f))一下，本来以为开销不大，没想到大数据一直超时，始终找不到问题，

想想算法没错呀，O(n^2)应该是最低了，没想到是memset搞得鬼，以后真是不能随便一个memset了。

class Solution {
public:
    int minDistance(string word1, string word2) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int> > f(word1.size()+1, vector<int>(word2.size()+1));
        
        f[0][0] = 0;
        for(int i = 1; i <= word2.size(); i++)
            f[0][i] = i;
        
        for(int i = 1; i <= word1.size(); i++)
            f[i][0] = i;
            
        for(int i = 1; i <= word1.size(); i++)
            for(int j = 1; j <= word2.size(); j++)
            {
                f[i][j] = INT_MAX;
                if (word1[i-1] == word2[j-1]) 
                    f[i][j] = f[i-1][j-1];
                
                f[i][j] = min(f[i][j], f[i-1][j-1] + 1); //replace
                f[i][j] = min(f[i][j], min(f[i-1][j], f[i][j-1]) + 1); //delete or insert               
            }
            
        return f[word1.size()][word2.size()];
    }
};