leetcode 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

题目来源：https://leetcode.com/problems/edit-distance/

分析

动态规划。既然允许三种操作：插入、删除和替换，那么就三个操作都做然后选取一种最小的。dp[i][j]表示word1前i个字符和word2前j个字符之间的最小距离，递推公式就是：

如果word1的第i个字符和word2的第j个字符相同；
dp[i][j] = dp[i-1][j-1]
否则
dp[i][j] = min(dp[i-1][j-1]+1, dp[i-1][j] + 1, dp[i][j-1] + 1)

dp[i-1][j-1]+1表示增加一次替换操作；
dp[i-1][j]+1表示word2执行insert操作或者word1执行delete操作；
dp[i][j-1]+1表示word1执行insert操作或者word2执行delete操作。

代码

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