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

Output:0

We are given a linked list which may or may not have a loop. A loop in a linked list means that if we keep doing a next, we will never reach null. Thus, we can be clear of a fact that if we reach null, then the list does not have a loop.

So how do we detect if there is a loop in the linked list. One method that immediately comes to our mind is that if a value is encountered twice, there may be a chance that the linked list has a loop. Thus, we can use Sets to keep track of all the values traversed.

**Algorithm**

- Allocate a
*Set*to store*ListNode*references. - Traverse the list, checking
*visited*for containment of the current node. - If the node has already been seen, then it is necessarily the entrance to the cycle.
- If any other node were the entrance to the cycle, then we would have already returned that node instead.
- Otherwise, the if condition will never be satisfied, and our function will return null.

The algorithm necessarily terminates for any list with a finite number of nodes, as the domain of input lists can be divided into two categories: cyclic and acyclic lists. An acyclic list resembles a null-terminated chain of nodes, while a cyclic list can be thought of as an acyclic list with the final null replaced by a reference to some previous node. If the while loop terminates, we return null, as we have traversed the entire list without encountering a duplicate reference. In this case, the list is acyclic. For a cyclic list, the while loop will never terminate, but at some point the if condition will be satisfied and cause the function to return.

// Template for a linked list node class ListNode { int value; ListNode next; } public class Solution { public ListNode detectCycle(ListNode head) { Set<ListNode> visited = new HashSet<ListNode>(); ListNode node = head; // Traverse the list while (node != null) { // If the node is already present in the Set // this is a loop and we need to return it. if (visited.contains(node)) { return node; } visited.add(node); node = node.next; } // No loop found return null; } }

*Time Complexity:* O(n)

*Space Complexity:* O(n)

A sample problem can be found here.

A method with constant space can be found here.