344. Reverse String 最基础的旋转字符串 class Solution { public: void reverseString(vector<char>& s) { if(s.empty()) return; ; ; while(start < end){ char tmp = s[end]; s[end] = s[start]; s[start] = tmp; start++; end--; } return; } }; 541. Reverse Strin…

problem 796. Rotate String solution1: class Solution { public: bool rotateString(string A, string B) { if(A.size()!=B.size()) return false; && B.size()==) return true;//errr... ; i<A.size(); ++i) { , i) == B) return true; } return false; } }; s…

Question 796. Rotate String Solution 题目大意:两个字符串匹配 思路:Brute Force Java实现: public boolean rotateString(String A, String B) { if (A.length() != B.length()) return false; if (A.length() == 0) return true; // Brute Force for (int i=0; i<A.length(); i++) {…

［抄题］: We are given two strings, A and B. A shift on A consists of taking string A and moving the leftmost character to the rightmost position. For example, if A = 'abcde', then it will be 'bcdea' after one shift on A. Return True if and only if A can…

We are given two strings, A and B. A shift on A consists of taking string A and moving the leftmost character to the rightmost position. For example, if A = 'abcde', then it will be 'bcdea' after one shift on A. Return True if and only if A can becom…

class Solution { public: bool rotateString(string A, string B) { if(A.length()==B.length()&&(A+A).find(B)!=string::npos) return true; return false; } }; 把俩A拼起来,能找到B,则可以通过循环右移将A转化为B…

We are given two strings, A and B. A shift on A consists of taking string A and moving the leftmost character to the rightmost position. For example, if A = 'abcde', then it will be 'bcdea' after one shift on A. Return True if and only if A can becom…

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/rotate-string/description/ 题目描述 We are given two strings, A and B. A shift on A consists of taking string A and moving the leftmost charac…

这是悦乐书的第317次更新,第338篇原创 在开始今天的算法题前,说几句,今天是世界读书日,推荐两本书给大家,<终身成长>和<禅与摩托车维修艺术>,值得好好阅读和反复阅读. 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第186题(顺位题号是796).给定两个字符串A和B,在A上进行移位操作,规则是将A最左边的字符移动到最右边去.例如,如果A ='abcde',那么在A上移位一次后,它将是'bcdea'.当且仅当A在A上移位一定次数后可以变为B时返回True.…

Given a string and an offset, rotate string by offset. (rotate from left to right) Example Given "abcdefg". offset=0 => "abcdefg" offset=1 => "gabcdef" offset=2 => "fgabcde" offset=3 => "efgabcd&quo…

Given a string and an offset, rotate string by offset. (rotate from left to right) Example Given "abcdefg" for offset=0, return "abcdefg" for offset=1, return "gabcdef" for offset=2, return "fgabcde" for offset=3, r…

8. Rotate String Description Given a string and an offset, rotate string by offset. (rotate from left to right) Example Given "abcdefg". offset=0 => "abcdefg" offset=1 => "gabcdef" offset=2 => "fgabcde" off…

Description Given a string and an offset, rotate string by offset. (rotate from left to right) Example Given "abcdefg". offset=0 => "abcdefg" offset=1 => "gabcdef" offset=2 => "fgabcde" offset=3 => "…

We are given two strings, A and B. A shift on A consists of taking string A and moving the leftmost character to the rightmost position. For example, if A = 'abcde', then it will be 'bcdea' after one shift on A. Return True if and only if A can becom…

We are given two strings, A and B. A shift on A consists of taking string A and moving the leftmost character to the rightmost position. For example, if A = 'abcde', then it will be 'bcdea' after one shift on A. Return True if and only if A can becom…

We are given two strings, A and B. A shift on A consists of taking string A and moving the leftmost character to the rightmost position. For example, if A = 'abcde', then it will be 'bcdea' after one shift on A. Return True if and only if A can becom…

1.题目描述 2.问题分析 直接旋转字符串A,然后做比较即可. 3.代码 bool rotateString(string A, string B) { if( A.size() != B.size() ) return false; if( A.empty() && B.empty() ) return true; ; while( i < A.size() ){ ] ; A += c; A.erase( A.begin() ); if( A == B ) return true;…

We are given two strings, A and B. A shift on A consists of taking string A and moving the leftmost character to the rightmost position. For example, if A = 'abcde', then it will be 'bcdea' after one shift on A. Return True if and only if A can becom…

We are given two strings, A and B. A shift on A consists of taking string A and moving the leftmost character to the rightmost position. For example, if A = 'abcde', then it will be 'bcdea' after one shift on A. Return True if and only if A can becom…

给定两个字符串, A 和 B. A 的旋转操作就是将 A 最左边的字符移动到最右边. 例如, 若 A = 'abcde',在移动一次之后结果就是'bcdea' .如果在若干次旋转操作之后,A 能变成B,那么返回True. 示例 1: 输入: A = 'abcde', B = 'cdeab' 输出: true 示例 2: 输入: A = 'abcde', B = 'abced' 输出: false 注意: A 和 B 长度不超过 100. class Solution { public: bool…

突然很想刷刷题,LeetCode是一个不错的选择,忽略了输入输出,更好的突出了算法,省去了不少时间. dalao们发现了任何错误,或是代码无法通过,或是有更好的解法,或是有任何疑问和建议的话,可以在对应的随笔下面评论区留言,我会及时处理,在此谢过了. 过程或许会很漫长,也很痛苦,慢慢来吧. 编号 题名 过题率 难度 1 Two Sum 0.376 Easy 2 Add Two Numbers 0.285 Medium 3 Longest Substring Without Repeating C…

645. Set Mismatch The set S originally contains numbers from 1 to n. But unfortunately, due to the data error, one of the numbers in the set got duplicated to another number in the set, which results in repetition of one number and loss of another nu…

All LeetCode Questions List(Part of Answers, still updating) 题目汇总及部分答案(持续更新中) Leetcode problems classified by company 题目按公司分类(Last updated: October 2, 2017) .   Top Interview Questions # Title Difficulty Acceptance 1 Two Sum Medium 17.70% 2 Add Two N…

Q1. Rotate String(796) We are given two strings, A and B. A shift on A consists of taking string A and moving the leftmost character to the rightmost position. For example, if A = 'abcde', then it will be 'bcdea' after one shift on A. Return True if…

请点击页面左上角 -> Fork me on Github 或直接访问本项目Github地址:LeetCode Solution by Swift    说明:题目中含有$符号则为付费题目. 如:[Swift]LeetCode156.二叉树的上下颠倒 $ Binary Tree Upside Down 请下拉滚动条查看最新 Weekly Contest!!! Swift LeetCode 目录 | Catalog 序        号 题名Title 难度     Difficulty  两数之…

Note: 后面数字n表明刷的第n + 1遍, 如果题目有**, 表明有待总结 Conclusion questions: [LeetCode] questions conclustion_BFS, DFS LeetCode questions conclustion_Path in Tree [LeetCode] questions conlusion_InOrder, PreOrder, PostOrder traversal [LeetCode] questions for Dynamic…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 把自己刷过的所有题目做一个整理,并且用简洁的语言概括了一下思路,汇总成了一个表格. 题目的排列顺序是按照先Easy再Medium再Hard排列的,暂时还没有把题目全部整理完成.后序我会把刷过的所有的题目都整理到这个文档里. 题目 难度 解法 题目地址 566. Reshape the Matrix Easy 变长数组,求余法,维护行列计算在新的数组中的位置 https://blog.c…

Yet Another Source Code for LintCode Current Status : 232AC / 289ALL in Language C++, Up to date (2016-02-10) For more problems and solutions, you can see my LintCode repository. I'll keep updating for full summary and better solutions. See cnblogs t…

一直直到bug-free.不能错任何一点. 思路不清晰:刷两天. 做错了,刷一天. 直到bug-free.高亮,标红. 185,OA(YAMAXUN)--- (1) findFirstDuplicate string in a list of string. import java.util.HashSet; import java.util.Set; public class Solution { public static void main(String[] args) { String[…

--------------------------------------------------------------- 本文使用方法:所有题目,只需要把标题输入lintcode就能找到.主要是简单的剖析思路以及不能bug-free的具体细节原因. ---------------------------------------------------------------- ------------------------------------------- 第九周:图和搜索. ---…