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

There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail's next pointer is connected to (0-indexed). It is -1 if there is no cycle. Note that pos is not passed as a parameter.

Do not modify the linked list.

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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.


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Constraints:


	The number of the nodes in the list is in the range [0, 104].
	-105 &lt;= Node.val &lt;= 105
	pos is -1 or a valid index in the linked-list.


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Follow up: Can you solve it using O(1) (i.e. constant) memory?

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if not head or not head.next:
            return None
        slow = fast = entry = head
        
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            
            if slow == fast:
                while slow != entry:
                    slow = slow.next
                    entry = entry.next
                return entry
        return None
public class Solution {
    public ListNode detectCycle(ListNode head) {
        if (head == null) return null;
        ListNode tortoise = head;
        ListNode hare = new ListNode();
        hare.next = head.next;
        while (hare != null && hare.next != null && hare != tortoise) {
            tortoise = tortoise.next;
            hare = hare.next.next;
        }
        if (hare == null || hare.next == null) return null;
        tortoise = head;
        while (tortoise != hare) {
            tortoise = tortoise.next;
            hare = hare.next;
        }
        return hare;
    }
}
class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        map<ListNode*, int> m;
        int index = 0;
        while (head)
        {
            if (m.find(head) == m.end())
                m[head] = index;
            else 
                return (head);
            head = head->next;
            index++;       
        }
        return (NULL);
    }
};
var detectCycle = function(head) {
    let slowRunner = head
    let fastRunner = head

    while(fastRunner) {
        fastRunner = fastRunner.next?.next
        slowRunner = slowRunner.next

        if (fastRunner === slowRunner) {
            fastRunner = slowRunner
            slowRunner = head

            while (slowRunner !== fastRunner) {
                slowRunner = slowRunner.next
                fastRunner = fastRunner.next
            }

            return slowRunner
        }
    }

    return null
};
Linked List Cycle II