LeetCode 之Edit Distance

/*动态规划法。用dp[i][j]表示word1[i...m],word[j...n]的距离（m，n分别表示其最后一个元素）。1）若word1[i] == word2[j]，则有dp[i][j] = dp[i+1][j+1];2）若word1[i] != word2[j], 则有dp[i][j] = min{    dp[i+1][j+1]+1（这种情况是用word1[i]代替word2[j]或用word2[j]代替word1[i]），    dp[i][j+1]+1(这种情况是删掉word2[j]或者在word1[i]之前插入word2[j])    dp[i+1][j]+1(这种情况是删掉word1[i]或者在word2[j]之前插入word1[i])    }最后返回dp[0][0].时间空间复杂度都是O(n^2).*/class Solution {public:    int minDistance(string word1, string word2) {        if(word1.empty()) return word2.size();        if(word2.empty()) return word1.size();        vector<vector<int> > dp(word1.size()+1, vector<int>(word2.size()+1, 0));        for(int i = 0; i < word2.size()+1; ++i){            dp[word1.size()][i] = word2.size() - i;        }        for(int i = 0; i < word1.size()+1; ++i){            dp[i][word2.size()] = word1.size() - i;        }        for(int i = word1.size()-1; i >= 0; --i){            for(int j = word2.size()-1; j >= 0; --j){                if(word1[i] == word2[j]) dp[i][j] = dp[i+1][j+1];                else{                    int tmp = min(dp[i+1][j]+1, dp[i][j+1]+1);                    dp[i][j] = min(tmp, dp[i+1][j+1]+1);                 }            }        }        return dp[0][0];    }};