leetcode 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：利用动态规划的思路。dp[i][j]表示word1的前i个字母与word2的前j个字母的编辑距离。

总结的递推公式就是：

EDIT[i,j]表示对于字符串a从1到i的子串和字符串b从1到j的字串的编辑距离。（字符串下标从1开始)

EDIT[i - 1,j] + 1表示对a 在i 位置删除delete操作

EDIT[i,j - 1] + 1 表示insert添加操作

EDIT[i-1, j - 1] + f(x[i],y[j])这里如果x[i] == y[j] 则 f(x[i],y[j]) == 0 否则 ==1

表示不变或者是modify操作。

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