Question:

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.

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.

Constraints:

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

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

Solution:

• One approach is to use a HashSet to store all the nodes. If any such node is encountered while traversing that is already present in HashSet, that will be the required Node.
• Better space-optimized solution would be to use Floyd's Cycle Detection Algorithm.
• We use slow and fast pointers. If fast pointer reaches null at any point, then it means that there is no cycle. Otherwise, there is a cycle, and slow and fast pointers will meet at some node while traveling within the cycle.
• To identify the node at which the cycle starts, we reset the slow pointer to head. Now, we increment both the fast and slow pointers by 1, the point at which they will meet, will be the start node of the cycle.
• There is actually a proof available for the previous step that one can look up.

Code:

public class Solution {
    public ListNode detectCycle(ListNode head) {

        if(head == null || head.next == null) return null;

        ListNode slow = head, fast = head;

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

            slow = slow.next;
            fast = fast.next.next;

            if(fast == slow) break;

        }

        if(slow != fast) return null; 

        slow = head;

        while(fast != slow){

            slow = slow.next;
            fast = fast.next;

        }

        return slow;

    }
}

Complexity:

• Time: O(N)
• Space: O(1)