Edit Distance

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：典型的dp问题，用dp[i][j]表示word1[0:i)到word2[0:j)的编辑距离，递归思路如下：
if(word1[i-1] == word2[j-1]) dp[i][j] = dp[i-1][j-1]
else dp[i][j] = min(dp[i][j-1], dp[i-1][j-1], dp[i-1][j]) + 1
代码如下：

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

20ms通过