Given a linked list, determine if it has a cycle in it.
Follow up:
Can you solve it without using extra space?
检测链表中是否有环。
思路：
经典算法，使用快慢指针，快的一次走两步，慢的一次走一步，如果有环它们一定会相遇。
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 bool HasCycle(ListNode head) {
if(head == null){
return false;
}
var p = head;
var q = head;
while(p != null && q != null && q.next != null){
var t = q;
p = p.next;
q = q.next.next;
if(ReferenceEquals(p,q)){
return true;
}
}
return false;
}
}