题意:给一个单链表,判断其是否出现环! 思路:搞两个指针,每次,一个走两步,另一个走一步.若有环,他们会相遇,若无环,走两步的指针必定会先遇到NULL. /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: bool has…

Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? 这道题是快慢指针的经典应用.只需要设两个指针,一个每次走一步的慢指针和一个每次走两步的快指针,如果链表里有环的话,两个指针最终肯定会相遇.实在是太巧妙了,要是我肯定想不出来.代码如下: C++ 解法: class Solution { public: bool hasCycle…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up:Can you solve it without using extra space? Hide Tags Linked List Two Pointers        开始犯二了一点,使用了node 的val 作为判断,其实是根据内存判断.找出链表环的起始位置,这个画一下慢慢找下规律…

2.6 Given a circular linked list, implement an algorithm which returns the node at the beginning of the loop.DEFINITIONCircular linked list: A (corrupt) linked list in which a node's next pointer points to an earlier node, so as to make a loop in the…

Given a linked list, determine if it has a cycle in it. To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in…

Given a linked list, determine if it has a cycle in it. ExampleGiven -21->10->4->5, tail connects to node index 1, return true Challenge Follow up:Can you solve it without using extra space? LeetCode上的原题,请参见我之前的博客Linked List Cycle. class Solution…

Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 判断链表中是否有环,不能用额外的空间,可以使用快慢指针,慢指针一次走一步,快指针一次走两步,若是有环则快慢指针会相遇,若是fast->next==NULL则没有环. 值得注意的是:在链表的题中,快慢指针的使用频率还是很高,值得注意. /** * Definition for si…

Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 参考博客:链表环状检测 package cn.magicdu; import cn.magicdu.extra.ListNode; public class _114_Linked_List_Cycle { public boolean hasCycle(ListNode hea…

参考资料 • Floyd判圈算法 { 链接 } • 单链表“环”.“环的起点”.环的长度”问题 { 链接 } 链表环的问题 一.判断链表有换 使用两个指针slow和fast.两个指针开始时均在头节点处(SS点),slow指针(龟)一次向后移动一个一步,fast指针(兔)一次向后移动两步.若存在环,则slow和fast必能相遇:反之若slow到达链表尾时两个指针仍不能相遇,则不存在环. 代码:链接 如果我们把环从P点“切开”(当然并不是真的切,那就破坏原来的数据结构了),那么问题就转化为计算两个相…

Linked List Cycle II Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up: Can you solve it without using extra space? 和问题一Linked List Cycle几乎一样.如果用我的之前的解法的话,可以很小修改就可以实现这道算法了.但是如果问题一用优化了的解法的话,那么就不适…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up: Can you solve it without using extra space? 这个求单链表中的环的起始点是之前那个判断单链表中是否有环的延伸,可参见我之前的一篇文章 (http://www.cnblogs.com/grandyang/p/4137187.html). 还是要设…

题目要求 Linked List Cycle Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? 如何判断一个单链表中有环? Linked List Cycle II Given a linked list, return the node where the cycle begins. If there is no cycle…

题意:给一个单链表,若其有环,返回环的开始处指针,若无环返回NULL. 思路: (1)依然用两个指针的追赶来判断是否有环.在确定有环了之后,指针1跑的路程是指针2的一半,而且他们曾经跑过一段重叠的路(即1跑过,2也跑过),就是那段(环开始处,相遇处),那么指针2开始到环开始处的距离与head到指针相遇处是等长的喔~,那么再跑一次每次一步的就必定会相遇啦.画个图图好方便看~ (2)其实还有另一个直观的思路,就是指针1和2相遇后,p指向他们的next,在他们相遇处的next给置空,再跑一遍那个“找两…

Given a linked list, return the node where the cycle begins. If there is no cycle, returnnull. Follow up:Can you solve it without using extra space? 题意:给定链表,若是有环,则返回环开始的节点,没有则返回NULL 思路:题目分两步走,第一.判断是否有环,第二若是有,找到环的起始点.关于第一点,可以参考之前的博客 Linked list cycle.…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up:Can you solve it without using extra space? 141. Linked List Cycle 的拓展,这题要返回环开始的节点,如果没有环返回null. 解法:双指针,还是用快慢两个指针,相遇时记下节点.参考:willduan的博客 Java: pu…

描述 给定一个链表,判断它是否有环. 样例 给出 -21->10->4->5, tail connects to node index 1,返回 true. 这里解释下,题目的意思,在英文原题中,tail connects to node index 1 表示的是节点 5 还要链接回索引号 为 1 的节点. 一个典型的带环链表如下: 挑战 不要使用额外的空间 代码 GitHub 的源代码,请访问下面的链接: https://github.com/cwiki-us/java-tutoria…

141. Linked List Cycle Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 利用快慢指针,如果相遇则证明有环 注意边界条件: 如果只有一个node. public class Solution { public boolean hasCycle(ListNode head) { if(head==null |…

本题要求将给定的单链表翻转,是校招面试手撕代码环节的高频题,能很好地考察对单链表这一最简单数据结构的理解:可以使用迭代和递归两种方法对一个给定的单链表进行翻转,具体实现如下: class Solution { public: ListNode* reverseList(ListNode* head) { return reverseList_iteratively(head); //return reverseList_recursively(head);//递归方法leetcode超时 } /…

Linked List Cycle Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? Linked List Two Pointers     ''' Created on Nov 13, 2014 @author: ScottGu<gu.kai.66@gmail.com, 150316990@qq.com> ''' # De…

一.判断链表是否存在环,办法为: 设置两个指针(fast, slow),初始值都指向头,slow每次前进一步,fast每次前进二步,如果链表存在环,则fast必定先进入环,而slow后进入环,两个指针必定相遇. 二.找到环的入口点 当fast若与slow相遇时,slow肯定没有走遍历完链表,而fast已经在环内循环了n圈(1<=n).假设slow走了s步,则fast走了2s步(fast步数还等于s 加上在环上多转的n圈),设环长为r,则: 2s = s + nrs= nr 设整个链表长L,入口环…

Linked List Cycle II Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up:Can you solve it without using extra space? SOLUTION 1: 1. 先用快慢指针判断是不是存在环. 2. 再把slow放回Start处,一起移动,直到二个节点相遇,就是交点.…

1.问题描述 Remove all elements from a linked list of integers that have value val. ExampleGiven: 1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6, val = 6Return: 1 --> 2 --> 3 --> 4 --> 5 给定一个单链表和一个数值,删除单链表中数据域等于该数值的节点. 2.问题分析 遍历一次链表,找到数据域等…

Linked List Cycle Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? SOLUTION 1: 经典快慢指针问题.如果存在环,fast, slow必然会相遇.就像2个速度不一样的人在环形跑道赛跑,总有一个时间他们会相遇. 如果fast到达了null,就是没有环,可以返回false. package Algorith…

Question : Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up: Can you solve it without using extra space? Anaylsis : 首先,比较直观的是,先使用Linked List Cycle I的办法,判断是否有cycle.如果有,则从头遍历节点,对于每一个节点,查询是否在环里面,是…

Linked List Cycle Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? 错误解法.因为Cycle可能出现在Link的中间的,所以需要检查其中间Nodes. bool hasCycle(ListNode *head) { // IMPORTANT: Please reset any member data you…

链表是有序的列表,但是在内存中存储图下图所示 链表是以 节点 的方式来存储,是 链式存储 每个节点包含 data 域.next 域,指向下一个节点 链表的各个节点 不一定是连续存储,如上图所示 链表还分:带头节点.不带头节点,根据实际需求来确定 上面进行了一个简单的介绍,下面分几部分来讲解: 目录 单链表 单链表的应用实例 单链表-无排序实现 单链表-有序实现(从小到大) 单链表的修改 单链表的删除 单链表面试题 求单链表中有效节点的个数 查找单链表中的倒数第 k 个结点 单链表的反转 从尾到头…

逆转单链表,比较简单,不细讲,扫描依次改变指针指向. class Solution { public: ListNode* reverseList(ListNode* head) { if(head==nullptr)return head; ListNode * tmp = head->next; ListNode *prenode = head; while(tmp!=NULL) { ListNode *nextnode = tmp->next; tmp->next = prenod…

Given a linked list, determine if it has a cycle in it. To represent a cycle in the given linked list, we use an integer poswhich represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in t…

题目: Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up: Can you solve it without using extra space? 题解: 这个连同I都是很经典的题啦,刷CC150时候就折磨了半天. 其实就推几个递推公式就好..首先看图(图引用自CC150): 从链表起始处到环入口长度为:a,从环入口到Faster和Sl…

Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 题意: 给定一个链表,判断是否有环 思路: 快慢指针 若有环,则快慢指针一定会在某个节点相遇(此处省略证明) 代码: public class Solution { public boolean hasCycle(ListNode head) { ListNode fast =…