Level: Medium 题目描述: 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 wh…
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 poswhich represents the position (0-indexed) in the linked list where tail connects to. I…
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…
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 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.…
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 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…
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…