1.     public ListNode detectCycle(ListNode head) {
    2.         HashSet<ListNode> set = new HashSet<ListNode>();
    3.         while (head != null) {
    4.             // 检查是否有重复添加
    5.             if (!set.add(head)) {
    6.                 return head;
    7.             }
    8.             head = head.next;
    9.         }
    10.         return null;
    11.     }
    1. public class Solution {
    2.     public ListNode detectCycle(ListNode head) {
    3.         ListNode slow = head;
    4.         ListNode fast = head;
    5.         while (fast != null && fast.next != null) {
    6.             slow = slow.next;
    7.             fast = fast.next.next;
    8.             if (slow == fast) {
    9. //                当快慢在点p处相遇时，它们运行的长度是“a + 2b + c”和“a + b”。
    10. //                因为快比慢快2倍。所以a+2b+c == 2（a+b），然后我们得到'a==c'。
    11.                 ListNode slow2 = head;
    12.                 while (slow != slow2) {
    13.                     slow = slow.next;
    14.                     slow2 = slow2.next;
    15.                 }
    16.                 return slow;
    17.             }
    18.         }
    19.         return null;
    20.     }
    21. }

    https://leetcode.com/problems/linked-list-cycle-ii/discuss/44774/Java-O(1)-space-solution-with-detailed-explanation.-space-solution-with-detailed-explanation.)