编辑距离

给出两个单词word1和word2，计算出将word1 转换为word2的最少操作次数。

你总共三种操作方法：

• 插入一个字符
• 删除一个字符
• 替换一个字符
  

样例

给出 work1="mart" 和 work2="karma"

返回 3

class Solution {public:/*** @param word1 & word2: Two string.* @return: The minimum number of steps.*/int minDistance(string word1, string word2) {// write your code hereint n = word1.length() + 1;int m = word2.length() + 1;int **arr;arr = new int*[n];for (int i = 0; i < n; i++){arr[i] = new int[m];}for (int i = 0; i < n; i++)arr[i][0] = i;for (int j = 0; j < m; j++)arr[0][j] = j;for (int i = 1; i < n; i++)for (int j = 1; j < m; j++){if (word1[i - 1] == word2[j - 1])arr[i][j] = arr[i - 1][j - 1];else    arr[i][j] = min   (arr[i - 1][j - 1] + 1,     //替换一个字符   arr[i][j - 1] + 1,          //插入一个字符   arr[i - 1][j] + 1);         //删除一个字符}return arr[n-1][m-1];}int min(int a, int b, int c){int temp = a > b ? b : a;return temp > c ? c : temp;}};