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 from
p = head;
while(!ReferenceEquals(p, q)){
p = p.next;
q = q.next;
}
return q;
}
}