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：

这个去年学算法的时候学过，当时觉得好难啊，现在一看，这题真简单。就是一道常规的动态规划题目。直接AC，高兴~~

用dp[m][n]存储word1的前m个字符与word2的前n个字符的最小匹配距离。

那么dp[i][j] = dp[i-1][j]+1 （word1删除一个字符）、dp[i][j-1] + 1 （word2删除一个字符）、 dp[i-1][j-1]  + ((word1[i-1]==word2[j-1]) ? 0（当前字符相等） : 1（替换）)) 中最小的值。

class Solution {public:    int minDistance(string word1, string word2) {        int len1 = word1.length();        int len2 = word2.length();                vector
     
      
       > dp(len1 + 1, vector
       
        (len2 + 1, 0));        for(int i = 1; i < len1 + 1; i++)        {            dp[i][0] = i;        }        for(int j = 1; j < len2 + 1; j++)        {            dp[0][j] = j;        }        for(int i = 1; i < len1 + 1; i++)        {            for(int j = 1;j < len2 + 1; j++)            {                dp[i][j] = min(min(dp[i-1][j] + 1, dp[i][j-1] + 1), dp[i-1][j-1] + ((word1[i-1]==word2[j-1]) ? 0 : 1));            }        }        return dp[len1][len2];    }};