【leetcode】Edit Distance (hard)

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<vector<int>> dp(len1 + , vector<int>(len2 + , ));
        for(int i = ; i < len1 + ; i++)
        {
            dp[i][] = i;
        }
        for(int j = ; j < len2 + ; j++)
        {
            dp[][j] = j;
        }
        for(int i = ; i < len1 + ; i++)
        {
            for(int j = ;j < len2 + ; j++)
            {
                dp[i][j] = min(min(dp[i-][j] + , dp[i][j-] + ), dp[i-][j-] + ((word1[i-]==word2[j-]) ?  : ));
            }
        }
        return dp[len1][len2];
    }
};