> For the complete documentation index, see [llms.txt](https://jimmylin1991.gitbook.io/practice-of-algorithm-problems/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://jimmylin1991.gitbook.io/practice-of-algorithm-problems/linked-list/142.-linked-list-cycle-ii.md).

# 142. 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.
```

![](https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist.png)

**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.
```

![](https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist_test2.png)

**Example 3:**

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

![](https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist_test3.png)

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

```cpp
// Slow and Fast Pointers
ListNode *detectCycle(ListNode *head) { // time: O(n); space: O(1)
    ListNode *slow = head, *fast = head;
    while (fast && fast->next) {
        slow = slow->next;
        fast = fast->next->next;
        if (slow == fast) break;
    }
    if (!fast || !fast->next) return nullptr;
    slow = head;
    while (slow != fast) {
        slow = slow->next;
        fast = fast->next;
    }
    return slow;
}
```