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

思路：普通的DP很好写，问题是路径压缩的DP压缩如果进行，空间如何重复利用，和背包问题有点像，留待第二遍再解决

class Solution {public:    int minDistance(string word1, string word2) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int ** dp = new int*[word1.size() + 1];        for(int i = 0; i < word1.size() + 1; i++){            dp[i] = new int[word2.size() + 1];        }                for(int i = 0; i < word1.size() + 1; i++){            dp[i][0] = i;        }        for(int j = 0; j < word2.size() + 1; j++){            dp[0][j] = j;        }                for(int i = 1; i < word1.size() + 1; i++){            for(int j = 1; j < word2.size() + 1; j++){                if (word1[i-1] == word2[j-1]){                    //相等的情况下，最后一个元素不占用操作                    dp[i][j] = dp[i-1][j-1];                }else{                    //delete一种串                    dp[i][j] = min(dp[i-1][j] + 1,dp[i][j-1] + 1);                    //i和j的位置进行replace                    dp[i][j] = min(dp[i][j],dp[i-1][j-1] + 1);                }            }        }                    int ans = dp[word1.size()][word2.size()];        for(int i = 0; i < word1.size() + 1; i++){            delete []dp[i];        }        delete []dp;        return ans;    }};