[LeetCode]72.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

具体参考：[经典面试题]字符串编辑距离

思路一 超时

代码

/*——————————————–* 日期：2014-03-01* 作者：SJF0115* 题目: 72.Edit Distance* 网址：https://oj.leetcode.com/problems/edit-distance/* 结果：AC* 来源：LeetCode* 博客：————————————————*/;class Solution {public:int minDistance(string word1, string word2) {int m = word1.size();int n = word2.size();// Edit[i][j]为word1[0..i-1]和word2[0..j-1]的最小编辑数int Edit[m+1][n+1];// 初始化for(int i = 0;i <= m;++i){Edit[i][0] = i;}//forfor(int i = 0;i <= n;++i){Edit[0][i] = i;}//forfor(int i = 1;i <= m;++i){for(int j = 1;j <= n;++j){// 当前字符相同if(word1[i-1] == word2[j-1]){Edit[i][j] = Edit[i-1][j-1];}//ifelse{Edit[i][j] = 1 + min(Edit[i-1][j-1],min(Edit[i-1][j],Edit[i][j-1]));}//else}//for}//forreturn Edit[m][n];}};int main(){Solution solution;string str1(“ab”);string str2(“bc”);cout<<solution.minDistance(str1,str2)<<endl;return 0;}

运行时间

，喜欢真实的人，要做真实的人，所以从来不会想要刻意模仿任何人。