题目要求 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? 如何找到环的第一个节点? 根据题目意图,我们可以构建如下模型: 设:链表头是X,环的第一个节点是Y,slow和fast第一次的交点是Z.各段的长度分别是a,b,c,如图所示.环的长度是L.s…

题目描述: 中文: 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. 英文: Given a linked list, return the node where the cycle begins. If there is no cycle, return null. To represent…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. 解题: 寻找单链表中环的入口,如果不存在环则返回null. 假设单链表有环,先利用判断单链表是否有环(Linked List Cycle)的方法,找到快慢指针的交点. 假设环的入口在第K个结点处,那么此时快慢指针的交点,距离环入口,还差k个距离. 快慢指针相交后,在链表头位置同时启动另一个指针target,…

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…

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…

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…

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…

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…

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…

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…

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…

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…

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…

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)…

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…

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…

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? Subscribe to see which companies asked this question 查看是否有环,快慢两…

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.…

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…

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…

[114-Flatten Binary Tree to Linked List(二叉树转单链表)] [LeetCode-面试算法经典-Java实现][全部题目文件夹索引] 原题 Given a binary tree, flatten it to a linked list in-place. For example, Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 题目…

/*寻找单链表中数据域大小为k的结点,并与前一结点交换,如果前一结点存在的情况下*/ /* 算法思想:定义两个指针,pre指向前驱结点,p指向当前结点,当p->data == k的时候,交换 pre->data和p->data */ void SwapData(LinkList& L, int k) { LNode *pre = L, *p = L->next; int temp; while (p) { if (p->data == k) { temp = p-&g…

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, return null. 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.…

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? 思路 这题是Linked List Cycle的进阶版 Given a linked list, determine if it has a cycle in it. bool hasCycle(Li…