题目描述：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?判断链表是否有环，如果存在，返回环起始节点；如果不存在，返回Null。思路：1. 使用快慢指针的方法找到环的位置。2. 如果找到了环，慢指针回到起点，快慢指针每次各走一步，下一次相遇的位置就是环的起点。实现代码：/** * Definition for singly-linked list. * public class ListNode { *public int val; *public ListNode next; *public ListNode(int x) { *val = x; *next = null; *} * } */public class Solution {public ListNode DetectCycle(ListNode head){if(head == null){return null;}var p = head;var q = head;var found = false;while(p != null && q != null && q.next != null && !found){var t = q;p = p.next;q = q.next.next;if(ReferenceEquals(p,q)){found = true;}}if(!found){return null;}// p start from head again// and q standing where it is// next time they meet point is where cycle starts fromp = head;while(!ReferenceEquals(p, q)){p = p.next;q = q.next;}return q;}}
生活若剥去了理想梦想幻想，那生命便只是一堆空架子
