Edit Distance

题目：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：

f[i][j]代表word1[0...i]->word2[0...j]的步数;

f[i-1][j]      --  删除操作

f[i-1][j-1]  --  相等操作

f[i-1][j-1] +1 --  替换操作

f[i][j-1]  --  插入操作

代码：

class Solution {public://https://leetcode.com/problems/edit-distance///http://blog.163.com/gjx_12358@126/blog/static/895363452014232191498///讲解非常详细。    int minDistance(string word1, string word2) {        int row=word1.length()+1;        int col=word2.length()+1;        //为了初始化，从0到某个值一直就为某个值，见下面的初始化        vector<vector <int> > f(row,vector<int>(col));        //vector<vector <int> > f(row,vector<int>(col));                for(int i=0;i<row;i++)            f[i][0]=i;        for(int i=0;i<col;i++)            f[0][i]=i;        //初始化完毕                for(int i=1;i<row;i++){            for(int j=1;j<col;j++){                //如果相等                if(word1[i-1]==word2[j-1])                    f[i][j]=f[i-1][j-1];                else{                    f[i][j]=f[i-1][j-1]+1;//替换的情况                }                //真正开始比较的时候                f[i][j]=min(f[i][j],min(f[i][j-1]+1,f[i-1][j]+1));                           //删除          插入       刚刚的替换                }        }        return f[row-1][col-1];    }};