题目描述 给定两个字符串,你需要从这两个字符串中找出最长的特殊序列.最长特殊序列定义如下:该序列为某字符串独有的最长子序列(即不能是其他字符串的子序列). 子序列可以通过删去字符串中的某些字符实现,但不能改变剩余字符的相对顺序.空序列为所有字符串的子序列,任何字符串为其自身的子序列. 输入为两个字符串,输出最长特殊序列的长度.如果不存在,则返回 -1. 示例 : 输入: "aba", "cdc" 输出: 3 解析: 最长特殊序列可为 "aba"…

521. Longest Uncommon Subsequence I[easy] Given a group of two strings, you need to find the longest uncommon subsequence of this group of two strings. The longest uncommon subsequence is defined as the longest subsequence of one of these strings and…

problem 521. Longest Uncommon Subsequence I 最长非共同子序列之一 题意: 两个字符串的情况很少,如果两个字符串相等,那么一定没有非共同子序列,反之,如果两个字符串不等,那么较长的那个字符串就是最长非共同子序列. solution: class Solution { public: int findLUSlength(string a, string b) { ; return max(a.size(), b.size()); } }; or class…

Question 521. Longest Uncommon Subsequence I Solution 题目大意:给两个字符串,找出非共同子串的最大长度 思路:字符串相等就返回-1,不等就返回长度大的那个长度 Java实现: public int findLUSlength(String a, String b) { int aLength = a.length(); int bLength = b.length(); if (aLength != bLength) return Math.…

题目要求 Given a group of two strings, you need to find the longest uncommon subsequence of this group of two strings. The longest uncommon subsequence is defined as the longest subsequence of one of these strings and this subsequence should not be anysu…

Given a group of two strings, you need to find the longest uncommon subsequence of this group of two strings. The longest uncommon subsequence is defined as the longest subsequence of one of these strings and this subsequence should not be any subseq…

题目: anysubsequence of the other strings. A subsequence is a sequence that can be derived from one sequence by deleting some characters without changing the order of the remaining elements. Trivially, any string is a subsequence of itself and an empty…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.com/problems/longest-uncommon-subsequence-i/description/ 题目描述 Given a group of two strings, you need to find the longest uncommon subsequence of this g…

static int wing=[]() { std::ios::sync_with_stdio(false); cin.tie(NULL); ; }(); class Solution { public: int findLUSlength(string a, string b) { :max(a.length(),b.length()); } }; 骚题目,看看就好,认真你就输了…

［抄题］: [暴力解法]: 时间分析: 空间分析: [优化后]: 时间分析: 空间分析: ［奇葩输出条件］: ［奇葩corner case］: ［思维问题］: ［一句话思路］: 两个单词的话,就是看谁长 ［输入量］:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入): ［画图］: ［一刷］: ［二刷］: ［三刷］: ［四刷］: ［五刷］: [五分钟肉眼debug的结果]: ［总结］: ［复杂度］:Time complexity: O() Space complex…