大意：给定两个字符串word1和word2，为了使word1变为word2，可以进行增加、删除、替换字符三种操作，请输出操作的最少次数

 

Example 1:

Input: word1 = "horse", word2 = "ros"Output: 3Explanation: horse -> rorse (replace 'h' with 'r')rorse -> rose (remove 'r')rose -> ros (remove 'e')

Example 2:

Input: word1 = "intention", word2 = "execution"Output: 5Explanation: intention -> inention (remove 't')inention -> enention (replace 'i' with 'e')enention -> exention (replace 'n' with 'x')exention -> exection (replace 'n' with 'c')exection -> execution (insert 'u')

 

状态：dp[i][j]把word1[0..i-1]转换到word2[0..j-1]的最少操作次数

状态转移方程：

　　(1)如果word1[i-1] == word2[j-1]，则令dp[i][j] = dp[i-1][j-1]

　　(2)如果word1[i-1] != word2[j-1]，由于没有一个特别有规律的方法来断定执行何种操作，在增加、删除、替换三种操作中选一种操作次数少的赋值给dp[i][j];

　　　 增加操作：dp[i][j] = dp[i][j-1] + 1

　　　 删除操作：dp[i][j] = dp[i-1][j] + 1

　　     替换操作：dp[i][j] = dp[i-1][j-1] + 1

 

1 int minDistance(string word1,string word2){ 2     int wlen1 = word1.size(); 3     int wlen2 = word2.size(); 4      5     int**dp = new int*[wlen1 + 1]; 6     for (int i = 0; i <= wlen1; i++) 7         dp[i] = new int[wlen2 + 1]; 8      9     //int dp[maxn][maxn] = { 0 };10     for (int i = 0; i <= wlen1; i++)11         dp[i][0] = i;12     for (int j = 0; j <= wlen2; j++)13         dp[0][j] = j;14     int temp = 0;15     for (int i = 1; i <= wlen1; i++){16         for (int j = 1; j <= wlen2; j++){17             if (word1[i - 1] == word2[j - 1])dp[i][j] = dp[i - 1][j-1];18             else{19                 temp = dp[i - 1][j - 1]
      
       < dp[i][j - 1] ? temp : dp[i][j - 1];21                 dp[i][j] = temp + 1;22             }23         }24     }25 26     /*27     for (int i = 0; i <= wlen1; i++)28         delete[]dp[i];29     delete[]dp;30     */31 32     return dp[wlen1][wlen2];33 }