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? 头结点到cycle begins的点 距离是A, cycle begins的点 到快慢结点相遇的 点的距离是B A+B+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? 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…

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…

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? 快慢指针找重合判断是否循环.而后fast从头开始+1,slow继续前进直到重合 思路解析: 首先我们看一张图; 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…

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…

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…

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…