LeetCode 72. 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

分析：假设dp[i][j]表示word1[0…i]和word2[0…j]的最小编辑距离。
如果word1为空，则删除word2中元素
如果word2为空，则删除word1中元素
都不为空，有三种选择：
1）删除word1的最后一个字符，此时dp[i-1][j] + 1
2) 删除word2的最后一个字符，此时dp[i][j-1] + 1
3) word1最后一个替换为word2最后一个 , 则dp[i-1][j-1] + (word1[i] != word2[j])(下标从1开始)

代码如下：

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