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：动态规划法。
这是一个很好的二维动态规划问题，状态之间只存在一步转换，考虑dp[i][j]表示长度为i的word1转换到长度为j的word2话，有两种情况：
word1[i-1]==word2[j-1],dp[i][j]=dp[i-1][j-1]；
word1[i-1]！=word2[j-1]，min(dp[i-1][j],dp[i][j-1],dp[i-1][j-1])+1.
注意初始化就是一直插入和删除就好了。

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