题解

利用快慢指针相遇

/**
 * 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 ||
           head.next == null ||
           head.next.next == null) {
            return false;
        }
        ListNode slow = head;
        ListNode fast = head;
        do {
            slow = slow.next;
            fast = fast.next.next;
        }
        while (fast != null && fast.next != null && slow != fast);

        if (fast == null || fast.next == null) {
            return false;
        }

        return true;
    }
}