Linked List Cycle II

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

Note: Do not modify the linked list.

Follow up:

Can you solve it without using extra space?

Solution:

  1. /**
  2. * Definition for singly-linked list.
  3. * class ListNode {
  4. * int val;
  5. * ListNode next;
  6. * ListNode(int x) {
  7. * val = x;
  8. * next = null;
  9. * }
  10. * }
  11. */
  12. public class Solution {
  13. public ListNode detectCycle(ListNode head) {
  14. ListNode slow = head, fast = head;
  15. while (fast != null && fast.next != null) {
  16. slow = slow.next;
  17. fast = fast.next.next;
  18. if (slow == fast)
  19. break;
  20. }
  21. if (fast == null || fast.next == null)
  22. return null;
  23. slow = head;
  24. while (slow != fast) {
  25. slow = slow.next;
  26. fast = fast.next;
  27. }
  28. return slow;
  29. }
  30. }