Linked List Cycle II

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

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Note: Do not modify the linked list.

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.

Follow-up: Can you solve it without using extra space?

public class Solution {
    public ListNode checkLoop(ListNode head) {
        ListNode slow = head, fast = head;
        while (slow != null && fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            if (slow == fast)
                return fast;
        }
        return null;
    }

    public ListNode detectCycle(ListNode head) {
        if (head == null || head.next == null)
            return null;
        ListNode cycle = checkLoop(head);
        if (cycle == null)
            return cycle;
        ListNode temp = head;
        while (cycle != temp) {
            cycle = cycle.next;
            temp = temp.next;
        }
        return cycle;
    }
}