leetcode hot 20

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;
    }
}