题意: 给定两个字符串 A 和 B,求最长公共子串. 分析: 字符串的任何一个子串都是这个字符串的某个后缀的前缀. 求 A 和 B 的最长公共子串等价于求 A 的后缀和 B 的后缀的最长公共前缀的最大值.如果枚举 A和 B 的所有的后缀,那么这样做显然效率低下. 由于要计算 A 的后缀和 B 的后缀的最长公共前缀,所以先将第二个字符串写在第一个字符串后面,中间用一个没有出现过的字符隔开,再求这个新的字符串的后缀数组. 观察一下,看看能不能从这个新的字符串的后缀数组中找到一些规律.以 A=“aaa
题目链接:https://vjudge.net/problem/POJ-3450 Corporate Identity Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 8046 Accepted: 2710 Description Beside other services, ACM helps companies to clearly state their “corporate identity”, which i
题目链接:https://vjudge.net/problem/POJ-3294 Life Forms Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 16905 Accepted: 4970 Description You may have wondered why most extraterrestrial life forms resemble humans, differing by superficial t
Simpsons' Hidden Talents Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4597 Accepted Submission(s): 1671 Problem Description Homer: Marge, I just figured out a way to discover some of the
题目描写叙述: Write a function to find the longest common prefix string amongst an array of strings.就是给定1个字符串数组,找出公共最长前缀. 思路非常直接.使用1个索引来存最长公共前缀的长度就能够了. 注意, 假设使用1个字符串变量来存前缀的话,是不能AC的,由于题目不同意使用额外的空间. public string LongestCommonPrefix(string[] strs) { if(strs
题目链接:https://vjudge.net/problem/POJ-3415 Common Substrings Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 12240 Accepted: 4144 Description A substring of a string T is defined as: T(i, k)=TiTi+1...Ti+k-1, 1≤i≤i+k-1≤|T|. Given two stri
Justice String Given two strings A and B, your task is to find a substring of A called justice string, which has the same length as B, and only has at most two characters different from B. Input The first line of the input contains a single integer
Write a function to find the longest common prefix string amongst an array of strings. If there is no common prefix, return an empty string "". Example 1: Input: ["flower","flow","flight"] Output: "fl" Exa
