# 142. Linked List Cycle II (Medium)

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 = , pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.
``` Can you solve it without using extra space?

## Solutions

``````/**
* Definition for singly-linked list.
* class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) {
*         val = x;
*         next = null;
*     }
* }
*/
public class Solution {
public ListNode detectCycle(ListNode head) {
ListNode slow = head;
ListNode fast = head;

while (fast!=null && fast.next!=null){
fast = fast.next.next;
slow = slow.next;

if (fast == slow){
ListNode slow2 = head;
while (slow2 != slow){
slow = slow.next;
slow2 = slow2.next;
}
return slow;
}
}
return null;
}
}
``````