Given a linked list, determine if it has a cycle in it.

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. If pos is -1, then there is no cycle in the linked list.

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.

判断链表是否有环：使用快慢指针去判断，如果有环，fast快指针先进入环，slow慢指针后进入，经过一定步的操作，快慢指针在环上相遇。
方法一：（C++）

 1 class Solution {
 2 public:
 3     bool hasCycle(ListNode *head) {
 4         ListNode *fast=head,*slow=head;
 5         while(fast&&fast->next){
 6             slow=slow->next;
 7             fast=fast->next->next;
 8             if(slow==fast)
 9                 return true;
10         }
11         return false;
12     }
13 };

方法二：（Java）

 1 public class Solution {
 2     public boolean hasCycle(ListNode head) {
 3         Set<ListNode> s=new HashSet();
 4         while(head!=null){
 5             if(s.contains(head))
 6                 return true;
 7             else
 8                 s.add(head);
 9             head=head.next;
10         }
11         return false;
12     }
13 }

Java集合中含有contains（）方法，判断其是否已经含有此节点，如果没有，则使用add()添加进去