解题思路:快慢指针，一个一步一步走，一个两步两步走，如果有环，则一定会相遇；如果没环，那么快的指针一定会找到null。
/*** Definition for singly-linked list.* class ListNode {* int val;* ListNode next;* ListNode(int x) {* val = x;* next = null;* }* }*/
public class Solution {public boolean hasCycle(ListNode head) {if(head==null) return false;ListNode dummy_head = new ListNode();dummy_head.next = head;ListNode fast = dummy_head.next.next;ListNode slow = dummy_head.next;while(fast!=slow){if(fast==null || fast.next==null || fast.next.next==null){return false;}fast = fast.next.next;slow = slow.next;}return true;}
}
