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? 思考:第一步:设环长为len,快慢指针q.p相遇时,q比p多走了k*len.        第二部:p先走k*len步,p,q一起走每次一步,相遇时p比q多走了一个环距离,此时q就是环开始结点. 通…

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…

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.…

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…

问题描述如下: 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? 从问题来看,如果可以充分利用额外空间的话,这个题目是不难的,然而题目提出了一个要求,能否在不使用任何额外空间的情况下解决这个问题. 通过反复思考,我觉得这题类似于追击问题,可以用一个…

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? 原题链接:https://oj.leetcode.com/problems/linked-list-cycle-ii/ 题目:给定一个链表.返回环開始的节点.如无环.返回null. public L…

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? 借用博客http://www.cnblogs.com/hiddenfox/p/3408931.html的图 设环的距离为L = (b+c),无环的距离为a, 假设在时间t相遇,慢指针行驶距离为x,则…

Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? 思考:快慢指针,快指针一次走两步,慢指针一次一步.若快指针跟慢指针指向同一个结点,则有环.若快指针到达链表末尾即指向NULL,说明没有环. /** * Definition for singly-linked list. * struct ListNode { * int va…

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? 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…