leetcode Edit Distance

作者：jostree 转载请注明出处 http://www.cnblogs.com/jostree/p/4051082.html

题目链接：leetcode Edit Distance

最短编辑距离，简单dp，dp[i,j]表示到str1的第i个字符和str2的第j个字符止的最短编辑距离。

dp[i,j] = min(dp[i-1,j-1]+ !(str1[i]==str2[j]), dp[i-1,j]+1, dp[i,j-1]+1)

注意二维动态vector的神奇定义方式：

vector<vector<int> > dp(word1.size()+1, vector<int> (word2.size()+1));

代码如下：

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