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…

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.…

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…

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…

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…

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…

题目: Given a linked list, return the node where the cycle begins. If there is no cycle, return null. 思路: 第一次相遇时slow走过的距离:a+b,fast走过的距离:a+b+c+b. 因为fast的速度是slow的两倍,所以fast走的距离是slow的两倍,有 2(a+b) = a+b+c+b,可以得到a=c(这个结论很重要!). 我们发现L=b+c=a+b,也就是说,从一开始到二者第一次相遇,…

# Definition for singly-linked list. # class ListNode(object): # def __init__(self, x): # self.val = x # self.next = None class Solution(object): def detectCycle(self, head): """ :type head: ListNode :rtype: ListNode """ if n…

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? 解题思路,本题和上题十分类似,但是需要观察出一个规律,参考LeetCode:Linked List Cycle II JAVA实现如下: public ListNode detectCycle(Li…

142. 环形链表 II 142. Linked List Cycle II 题目描述 给定一个链表,返回链表开始入环的第一个节点.如果链表无环,则返回 null. 为了表示给定链表中的环,我们使用整数 pos 来表示链表尾连接到链表中的位置(索引从 0 开始).如果 pos 是 -1,则在该链表中没有环. 说明: 不允许修改给定的链表. LeetCode142. Linked List Cycle II 示例 1: 输入: head = [3,2,0,-4], pos = 1 输出: tail…

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…

1.Linked List Cycle 题目链接 题目要求: Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space? 刚看到这道题,很容易写出下边的程序: bool hasCycle(ListNode *head) { ListNode *a = head, *b = head; while(a) { b = a->next; wh…

Linked List Cycle II 题解 题目来源:https://leetcode.com/problems/linked-list-cycle-ii/description/ Description Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Note: Do not modify the linked list. Follow up: C…

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…

LeetCode解题报告:Linked List Cycle && Linked List Cycle II 1题目 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 Cycle II Given a linked list, return the node w…

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 Cycle II Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Fol…

引入 快慢指针经常用于链表(linked list)中环(Cycle)相关的问题.LeetCode中对应题目分别是: 141. Linked List Cycle 判断linked list中是否有环 142. Linked List Cycle II 找到环的起始节点(entry node)位置. 简介 快指针(fast pointer)和慢指针(slow pointer)都从链表的head出发. slow pointer每次移动一格,而快指针每次移动两格. 如果快慢指针能相遇,则证明链表中有…

判断链表有环,环的入口结点,环的长度 1.判断有环: 快慢指针,一个移动一次,一个移动两次 2.环的入口结点: 相遇的结点不一定是入口节点,所以y表示入口节点到相遇节点的距离 n是环的个数 w + n + y = 2 (w + y) 经过化简,我们可以得到:w  = n - y; https://www.cnblogs.com/zhuzhenwei918/p/7491892.html 3.环的长度: 从入口结点或者相遇的结点移动到下一次再碰到这个结点计数 https://blog.csdn.ne…

1. Linked List Cycle II Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Note: Do not modify the linked list. 思路:想法是利用两指针,一个每次移动一步,另一个每次移动两步,如果存在环则这两个指针一定会相遇(这里可以在纸上画一下,因为后一个指针移动比前一个指针快,当后一个指针在环中来到前一个指针的…

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? 解法一: 使用unordered_map记录当前节点是否被访问过,如访问过返回该节点,如到达尾部说明无环. /** * Definition for sing…

题目: 141.Given a linked list, determine if it has a cycle in it. 142.Given a linked list, return the node where the cycle begins. If there is no cycle, return null. 思路: 带环链表如图所示.设置一个快指针和一个慢指针,快指针一次走两步,慢指针一次走一步.快指针先进入环,慢指针后进入环.在进入环后,可以理解为快指针追赶慢指针,由于两个指…

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? 解题思路: 判断链表有无环,可用快慢指针进行,快指针每次走两步,慢指针每次走一步,如果快指针追上了慢指针,则存在环,否则,快指针走到链表末尾即为NULL是也没追上,则无环. 为什么快慢指针可…

142. Linked List Cycle II[easy] Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Note: Do not modify the linked list. Follow up:Can you solve it without using extra space? 解法一: /** * Definition for singl…

链表相关题 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? (Easy) 分析: 采用快慢指针,一个走两步,一个走一步,快得能追上慢的说明有环,走到nullptr还没有相遇说明没有环. 代码: /** * Definition for singly-linked list. * s…

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? 说明:两个指针不同步长. /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x)…

一.Linked List Cycle Total Accepted: 85115 Total Submissions: 232388 Difficulty: Medium Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space?  (M) Linked List Cycle II   /** * Definition for sing…