Find first node of loop in a linked list

Write a function findFirstLoopNode() that checks whether a given Linked List contains a loop. If the loop is present then it returns point to the first node of the loop. Else it returns NULL.

Example : 

Input : Head of below linked list

Output : Pointer to node 2

We have discussed Floyd’s loop detection algorithm. Below are steps to find the first node of the loop.
1. If a loop is found, initialize a slow pointer to head, let fast pointer be at its position. 
2. Move both slow and fast pointers one node at a time. 
3. The point at which they meet is the start of the loop.

C++

// C++ program to return first node of loop.
#include <bits/stdc++.h>
using namespace std;

struct Node {
    int key;
    struct Node* next;
};

Node* newNode(int key)
{
    Node* temp = new Node;
    temp->key = key;
    temp->next = NULL;
    return temp;
}

// A utility function to print a linked list
void printList(Node* head)
{
    while (head != NULL) {
        cout << head->key << " ";
        head = head->next;
    }
    cout << endl;
}

// Function to detect and remove loop
// in a linked list that may contain loop
Node* detectAndRemoveLoop(Node* head)
{
    // If list is empty or has only one node
    // without loop
    if (head == NULL || head->next == NULL)
        return NULL;

    Node *slow = head, *fast = head;
    // Move slow and fast 1 and 2 steps
    // ahead respectively.
    slow = slow->next;
    fast = fast->next->next;
    // Search for loop using slow and
    // fast pointers
    while (fast && fast->next) {
        if (slow == fast)
            break;
        slow = slow->next;
        fast = fast->next->next;
    }
    // If loop does not exist
    if (slow != fast)
        return NULL;
    // If loop exists. Start slow from
    // head and fast from meeting point.
    slow = head;
    while (slow != fast) {
        slow = slow->next;
        fast = fast->next;
    }
    return slow;
}

/* Driver program to test above function*/
int main()
{
    Node* head = newNode(50);
    head->next = newNode(20);
    head->next->next = newNode(15);
    head->next->next->next = newNode(4);
    head->next->next->next->next = newNode(10);

    /* Create a loop for testing */
    head->next->next->next->next->next = head->next->next;

    Node* res = detectAndRemoveLoop(head);
    if (res == NULL)
        cout << "Loop does not exist";
    else
        cout << "Loop starting node is " << res->key;
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

C

// C++ program to return first node of loop.
#include <stdio.h>
#include <stdlib.h>

typedef struct Node {
    int key;
    struct Node* next;
} Node;

Node* newNode(int key)
{
    Node* temp = (Node*)malloc(sizeof(Node));
    temp->key = key;
    temp->next = NULL;
    return temp;
}

// A utility function to print a linked list
void printList(Node* head)
{
    while (head != NULL) {
        printf("%d  ",head->key);
        head = head->next;
    }
    printf("/n");
}

// Function to detect and remove loop
// in a linked list that may contain loop
Node* detectAndRemoveLoop(Node* head)
{
    // If list is empty or has only one node
    // without loop
    if (head == NULL || head->next == NULL)
        return NULL;

    Node *slow = head, *fast = head;

    // Move slow and fast 1 and 2 steps
    // ahead respectively.
    slow = slow->next;
    fast = fast->next->next;

    // Search for loop using slow and
    // fast pointers
    while (fast && fast->next) {
        if (slow == fast)
            break;
        slow = slow->next;
        fast = fast->next->next;
    }

    // If loop does not exist
    if (slow != fast)
        return NULL;

    // If loop exists. Start slow from
    // head and fast from meeting point.
    slow = head;
    while (slow != fast) {
        slow = slow->next;
        fast = fast->next;
    }

    return slow;
}

/* Driver program to test above function*/
int main()
{
    Node* head = newNode(50);
    head->next = newNode(20);
    head->next->next = newNode(15);
    head->next->next->next = newNode(4);
    head->next->next->next->next = newNode(10);

    /* Create a loop for testing */
    head->next->next->next->next->next = head->next->next;

    Node* res = detectAndRemoveLoop(head);
    if (res == NULL)
        printf("Loop does not exist");
    else
        printf("Loop starting node is %d ",res->key);

    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

// Java program to return
// first node of loop.
import java.util.*;
class GFG{

static class Node
{
  int key;
  Node next;
};

static Node newNode(int key)
{
  Node temp = new Node();
  temp.key = key;
  temp.next = null;
  return temp;
}

// A utility function to
// print a linked list
static void printList(Node head)
{
  while (head != null)
  {
    System.out.print(head.key + " ");
    head = head.next;
  }
  System.out.println();
}

// Function to detect and remove loop
// in a linked list that may contain loop
static Node detectAndRemoveLoop(Node head)
{
  // If list is empty or has
  // only one node without loop
  if (head == null || head.next == null)
    return null;

  Node slow = head, fast = head;

  // Move slow and fast 1
  // and 2 steps ahead
  // respectively.
  slow = slow.next;
  fast = fast.next.next;

  // Search for loop using
  // slow and fast pointers
  while (fast != null &&
         fast.next != null)
  {
    if (slow == fast)
      break;
    slow = slow.next;
    fast = fast.next.next;
  }

  // If loop does not exist
  if (slow != fast)
    return null;

  // If loop exists. Start slow from
  // head and fast from meeting point.
  slow = head;
  while (slow != fast)
  {
    slow = slow.next;
    fast = fast.next;
  }

  return slow;
}

// Driver code
public static void main(String[] args)
{
  Node head = newNode(50);
  head.next = newNode(20);
  head.next.next = newNode(15);
  head.next.next.next = newNode(4);
  head.next.next.next.next = newNode(10);

  // Create a loop for testing
  head.next.next.next.next.next = head.next.next;

  Node res = detectAndRemoveLoop(head);
  if (res == null)
    System.out.print("Loop does not exist");
  else
    System.out.print("Loop starting node is " +  res.key);
}
}

// This code is contributed by shikhasingrajput

Python3

# Python3 program to return first node of loop.
class Node:

    def __init__(self, key):

        self.key = key
        self.next = None

def newNode(key):

    temp = Node(key)
    return temp

# A utility function to print a linked list
def printList(head):

    while (head != None):
        print(head.key, end = ' ')
        head = head.next

    print()

# Function to detect and remove loop
# in a linked list that may contain loop
def detectAndRemoveLoop(head):

    # If list is empty or has only one node
    # without loop
    if (head == None or head.next == None):
        return None

    slow = head
    fast = head

    # Move slow and fast 1 and 2 steps
    # ahead respectively.
    slow = slow.next
    fast = fast.next.next

    # Search for loop using slow and
    # fast pointers
    while (fast and fast.next):
        if (slow == fast):
            break

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

    # If loop does not exist
    if (slow != fast):
        return None

    # If loop exists. Start slow from
    # head and fast from meeting point.
    slow = head

    while (slow != fast):
        slow = slow.next
        fast = fast.next

    return slow

# Driver code
if __name__=='__main__':

    head = newNode(50)
    head.next = newNode(20)
    head.next.next = newNode(15)
    head.next.next.next = newNode(4)
    head.next.next.next.next = newNode(10)

    # Create a loop for testing
    head.next.next.next.next.next = head.next.next

    res = detectAndRemoveLoop(head)

    if (res == None):
        print("Loop does not exist")
    else:
        print("Loop starting node is " +
              str(res.key))

# This code is contributed by rutvik_56

C#

// C# program to return
// first node of loop.
using System;
class GFG{

class Node
{
  public int key;
  public Node next;
};

static Node newNode(int key)
{
  Node temp = new Node();
  temp.key = key;
  temp.next = null;
  return temp;
}

// A utility function to
// print a linked list
static void printList(Node head)
{
  while (head != null)
  {
    Console.Write(head.key + " ");
    head = head.next;
  }
  Console.WriteLine();
}

// Function to detect and remove loop
// in a linked list that may contain loop
static Node detectAndRemoveLoop(Node head)
{
  // If list is empty or has
  // only one node without loop
  if (head == null || head.next == null)
    return null;

  Node slow = head, fast = head;

  // Move slow and fast 1
  // and 2 steps ahead
  // respectively.
  slow = slow.next;
  fast = fast.next.next;

  // Search for loop using
  // slow and fast pointers
  while (fast != null &&
         fast.next != null)
  {
    if (slow == fast)
      break;
    slow = slow.next;
    fast = fast.next.next;
  }

  // If loop does not exist
  if (slow != fast)
    return null;

  // If loop exists. Start slow from
  // head and fast from meeting point.
  slow = head;
  while (slow != fast)
  {
    slow = slow.next;
    fast = fast.next;
  }

  return slow;
}

// Driver code
public static void Main(String[] args)
{
  Node head = newNode(50);
  head.next = newNode(20);
  head.next.next = newNode(15);
  head.next.next.next = newNode(4);
  head.next.next.next.next = newNode(10);

  // Create a loop for testing
  head.next.next.next.next.next =
                      head.next.next;

  Node res = detectAndRemoveLoop(head);

  if (res == null)
    Console.Write("Loop does not exist");
  else
    Console.Write("Loop starting node is " +
                   res.key);
}
}

// This code is contributed by shikhasingrajput

Javascript

<script>
// Javascript program to return
// first node of loop.

class Node
{
    constructor(key)
    {
        this.key=key;
        this.next=null;
    }
}

// A utility function to
// print a linked list
function printList(head)
{
    while (head != null)
  {
    document.write(head.key + " ");
    head = head.next;
  }
  document.write("<br>");
}

// Function to detect and remove loop
// in a linked list that may contain loop
function detectAndRemoveLoop(head)
{
    // If list is empty or has
  // only one node without loop
  if (head == null || head.next == null)
    return null;

  let slow = head, fast = head;

  // Move slow and fast 1
  // and 2 steps ahead
  // respectively.
  slow = slow.next;
  fast = fast.next.next;

  // Search for loop using
  // slow and fast pointers
  while (fast != null &&
         fast.next != null)
  {
    if (slow == fast)
      break;
    slow = slow.next;
    fast = fast.next.next;
  }

  // If loop does not exist
  if (slow != fast)
    return null;

  // If loop exists. Start slow from
  // head and fast from meeting point.
  slow = head;
  while (slow != fast)
  {
    slow = slow.next;
    fast = fast.next;
  }

  return slow;
}

// Driver code
let head = new Node(50);
  head.next = new Node(20);
  head.next.next = new Node(15);
  head.next.next.next = new Node(4);
  head.next.next.next.next = new Node(10);

  // Create a loop for testing
  head.next.next.next.next.next = head.next.next;

  let res = detectAndRemoveLoop(head);
  if (res == null)
    document.write("Loop does not exist");
  else
    document.write("Loop starting node is " +  res.key);

// This code is contributed by unknown2108
</script>

Output:

Loop starting node is 15

Time Complexity: O(N), where N is length of array.
Auxiliary Space: O(1)

How does this approach work?
Let slow and fast meet at some point after Floyd’s Cycle finding algorithm. The below diagram shows the situation when the cycle is found.

We can conclude below from the above diagram 

Distance traveled by fast pointer = 2 * (Distance traveled 
                                         by slow pointer)

(m + n*x + k) = 2*(m + n*y + k)

Note that before meeting the point shown above, fast
was moving at twice speed.

x -->  Number of complete cyclic rounds made by 
       fast pointer before they meet first time

y -->  Number of complete cyclic rounds made by 
       slow pointer before they meet first time

From the above equation, we can conclude below 

    m + k = (x-2y)*n

Which means m+k is a multiple of n.

So if we start moving both pointers again at the same speed such that one pointer (say slow) begins from the head node of the linked list and other pointers (say fast) begins from the meeting point. When the slow pointer reaches the beginning of the loop (has made m steps), the fast pointer would have made also moved m steps as they are now moving the same pace. Since m+k is a multiple of n and fast starts from k, they would meet at the beginning. Can they meet before also? No, because the slow pointer enters the cycle first time after m steps.

Method 2:
In this method, a temporary node is created. The next pointer of each node that is traversed is made to point to this temporary node. This way we are using the next pointer of a node as a flag to indicate whether the node has been traversed or not. Every node is checked to see if the next is pointing to a temporary node or not. In the case of the first node of the loop, the second time we traverse it this condition will be true, hence we return that node. 
The code runs in O(n) time complexity and uses constant memory space.

Steps to solve this problem:

1. Create a temporary node and a pointer temp.

2. While head is not null:

    *Check if head->next is null then return null.

    *Check if head->next is temp than break.

    *Create a nex pointer to a node and point it to head->next.

    *Point head->next to temp.

    *Update head to nex.

3. Return head.

Below is the implementation of the above approach:

C++

// C++ program to return first node of loop
#include <bits/stdc++.h>
using namespace std;

struct Node {
    int key;
    struct Node* next;
};

Node* newNode(int key)
{
    Node* temp = new Node;
    temp->key = key;
    temp->next = NULL;
    return temp;
}

// A utility function to print a linked list
void printList(Node* head)
{
    while (head != NULL) {
        cout << head->key << " ";
        head = head->next;
    }
    cout << endl;
}

// Function to detect first node of loop
// in a linked list that may contain loop
Node* detectLoop(Node* head)
{

    // Create a temporary node
    Node* temp = new Node;
    while (head != NULL) {

        // This condition is for the case
        // when there is no loop
        if (head->next == NULL) {
            return NULL;
        }

        // Check if next is already
        // pointing to temp
        if (head->next == temp) {
            break;
        }

        // Store the pointer to the next node
        // in order to get to it in the next step
        Node* nex = head->next;

        // Make next point to temp
        head->next = temp;

        // Get to the next node in the list
        head = nex;
    }

    return head;
}

/* Driver program to test above function*/
int main()
{
    Node* head = newNode(50);
    head->next = newNode(20);
    head->next->next = newNode(15);
    head->next->next->next = newNode(4);
    head->next->next->next->next = newNode(10);

    /* Create a loop for testing */
    head->next->next->next->next->next = head->next->next;

    Node* res = detectLoop(head);
    if (res == NULL)
        cout << "Loop does not exist";
    else
        cout << "Loop starting node is " << res->key;

    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

C

// C program to return first node of loop
#include <stdio.h>
#include <stdlib.h>

typedef struct Node {
    int key;
    struct Node* next;
} Node;

Node* newNode(int key)
{
    Node* temp = (Node*)malloc(sizeof(Node));
    temp->key = key;
    temp->next = NULL;
    return temp;
}

// A utility function to print a linked list
void printList(Node* head)
{
    while (head != NULL) {
        printf("%d ", head->key);
        head = head->next;
    }
    printf("\n");
}

// Function to detect first node of loop
// in a linked list that may contain loop
Node* detectLoop(Node* head)
{

    // Create a temporary node
    Node* temp = (Node *)malloc(sizeof(Node));
    while (head != NULL) {
        // This condition is for the case when there is no
        // loop
        if (head->next == NULL)
            return NULL;
        // Check if next is already pointing to temp
        if (head->next == temp)
            break;
        // Store the pointer to the next node
        // in order to get to it in the next step
        Node* nex = head->next;
        // Make next point to temp
        head->next = temp;
        // Get to the next node in the list
        head = nex;
    }
    return head;
}

/* Driver program to test above function*/
int main()
{
    Node* head = newNode(50);
    head->next = newNode(20);
    head->next->next = newNode(15);
    head->next->next->next = newNode(4);
    head->next->next->next->next = newNode(10);

    /* Create a loop for testing */
    head->next->next->next->next->next = head->next->next;

    Node* res = detectLoop(head);
    if (res == NULL)
        printf("Loop does not exist");
    else
        printf("Loop starting node is %d ", res->key);

    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

// Java program to return first node of loop
import java.util.*;
class GFG{

static class Node
{
    int key;
    Node next;
};

static Node newNode(int key)
{
    Node temp = new Node();
    temp.key = key;
    temp.next = null;
    return temp;
}

// A utility function to print a linked list
static void printList(Node head)
{
    while (head != null)
    {
        System.out.print(head.key + " ");
        head = head.next;
    }
    System.out.println();
}

// Function to detect first node of loop
// in a linked list that may contain loop
static Node detectLoop(Node head)
{

    // Create a temporary node
    Node temp = new Node();
    while (head != null)
    {

        // This condition is for the case
        // when there is no loop
        if (head.next == null)
        {
            return null;
        }

        // Check if next is already
        // pointing to temp
        if (head.next == temp)
        {
            break;
        }

        // Store the pointer to the next node
        // in order to get to it in the next step
        Node nex = head.next;

        // Make next point to temp
        head.next = temp;

        // Get to the next node in the list
        head = nex;
    }

    return head;
}

/* Driver program to test above function*/
public static void main(String[] args)
{
    Node head = newNode(50);
    head.next = newNode(20);
    head.next.next = newNode(15);
    head.next.next.next = newNode(4);
    head.next.next.next.next = newNode(10);

    /* Create a loop for testing */
    head.next.next.next.next.next = head.next.next;

    Node res = detectLoop(head);
    if (res == null)
        System.out.print("Loop does not exist");
    else
        System.out.print("Loop starting node is " +
                                         res.key);

}
}

// This code is contributed by gauravrajput1

Python3

# Python3 program to return first node of loop
class Node:

    def __init__(self, x):

        self.key = x
        self.next = None

# A utility function to print a linked list
def printList(head):

    while (head != None):
        print(head.key, end = " ")
        head = head.next

# Function to detect first node of loop
# in a linked list that may contain loop
def detectLoop(head):

    # Create a temporary node
    temp = Node(-1)

    while (head != None):

        # This condition is for the case
        # when there is no loop
        if (head.next == None):
            return None

        # Check if next is already
        # pointing to temp
        if (head.next == temp):
            break

        # Store the pointer to the next node
        # in order to get to it in the next step
        nex = head.next

        # Make next point to temp
        head.next = temp

        # Get to the next node in the list
        head = nex

    return head

# Driver code
if __name__ == '__main__':

    head = Node(50)
    head.next = Node(20)
    head.next.next = Node(15)
    head.next.next.next = Node(4)
    head.next.next.next.next = Node(10)

    # Create a loop for testing
    head.next.next.next.next.next = head.next.next

    res = detectLoop(head)

    if (res == None):
        print("Loop does not exist")
    else:
        print("Loop starting node is ", res.key)

# This code is contributed by mohit kumar 29

C#

// C# program to return first node of loop
using System;

class GFG{

class Node
{
    public int key;
    public Node next;
};

static Node newNode(int key)
{
    Node temp = new Node();
    temp.key = key;
    temp.next = null;
    return temp;
}

// A utility function to print a linked list
static void printList(Node head)
{
    while (head != null)
    {
        Console.Write(head.key + " ");
        head = head.next;
    }
    Console.WriteLine();
}

// Function to detect first node of loop
// in a linked list that may contain loop
static Node detectLoop(Node head)
{

    // Create a temporary node
    Node temp = new Node();

    while (head != null)
    {

        // This condition is for the case
        // when there is no loop
        if (head.next == null)
        {
            return null;
        }

        // Check if next is already
        // pointing to temp
        if (head.next == temp)
        {
            break;
        }

        // Store the pointer to the next node
        // in order to get to it in the next step
        Node nex = head.next;

        // Make next point to temp
        head.next = temp;

        // Get to the next node in the list
        head = nex;
    }
    return head;
}

// Driver code
public static void Main(String[] args)
{
    Node head = newNode(50);
    head.next = newNode(20);
    head.next.next = newNode(15);
    head.next.next.next = newNode(4);
    head.next.next.next.next = newNode(10);

    // Create a loop for testing
    head.next.next.next.next.next = head.next.next;

    Node res = detectLoop(head);

    if (res == null)
        Console.Write("Loop does not exist");
    else
        Console.Write("Loop starting node is " +
                       res.key);
}
}

// This code is contributed by Amit Katiyar

Javascript

<script>
// javascript program to return first node of loop

     class Node {
        constructor() {
            this.key = 0;
            this.next = null;
        }
    }


    function newNode(key) {
         temp = new Node();
        temp.key = key;
        temp.next = null;
        return temp;
    }

    // A utility function to print a linked list
    function printList( head) {
        while (head != null) {
            document.write(head.key + " ");
            head = head.next;
        }
        document.write("<br/>");
    }

    // Function to detect first node of loop
    // in a linked list that may contain loop
    function detectLoop( head) {

        // Create a temporary node
         temp = new Node();
        while (head != null) {

            // This condition is for the case
            // when there is no loop
            if (head.next == null) {
                return null;
            }

            // Check if next is already
            // pointing to temp
            if (head.next == temp) {
                break;
            }

            // Store the pointer to the next node
            // in order to get to it in the next step
             nex = head.next;

            // Make next point to temp
            head.next = temp;

            // Get to the next node in the list
            head = nex;
        }

        return head;
    }

    /* Driver program to test above function */

         head = newNode(50);
        head.next = newNode(20);
        head.next.next = newNode(15);
        head.next.next.next = newNode(4);
        head.next.next.next.next = newNode(10);

        /* Create a loop for testing */
        head.next.next.next.next.next = head.next.next;

         res = detectLoop(head);
        if (res == null)
            document.write("Loop does not exist");
        else
            document.write("Loop starting node is " + res.key);


// This code contributed by gauravrajput1
</script>

Output:

Loop starting node is 15

Time Complexity: O(N)
Auxiliary Space: O(1)

Method 3:
We can also use the concept of hashing in order to detect the first node of the loop. The idea is simple just iterate over the entire linked list and store node addresses in a set(C++ STL) one by one, while adding the node address into the set check if it already contains that particular node address if not then add node address to set if it is already present in the set then the current node is the first node of the loop. 

C++14

// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002)
// The below function take head of Linked List as
// input and return Address of first node in
// the loop if present else return NULL.
#include<bits/stdc++.h>
using namespace std;
// Structure of linked list node
struct ListNode{
    int val;
    ListNode *next;
    ListNode(int x){
        this->val = x;
        this->next = NULL;
    }
};

ListNode* detectCycle(ListNode* A){
    // declaring map to store node address
    unordered_set<ListNode*> uset;
    ListNode *ptr = A;

    // Default consider that no cycle is present
    while (ptr != NULL) {
        // checking if address is already present in map
        if (uset.find(ptr) != uset.end())
          return ptr;
        // if address not present then insert into the set
        else
            uset.insert(ptr);

        ptr = ptr->next;
    }
    return NULL;
}

/* Driver program to test above function*/
int main(){
    ListNode* head = new ListNode(50);
    head->next = new ListNode(20);
    head->next->next = new ListNode(15);
    head->next->next->next = new ListNode(4);
    head->next->next->next->next = new ListNode(10);

    // creating a loop for linked list
    head->next->next->next->next->next = head->next->next;

    ListNode* res = detectCycle(head);
    if (res == NULL)
        cout << "Loop does not exist";
    else
        cout << "Loop starting node is " << res->val;

    return 0;
}

Java

// The below function take head of Linked List as
// input and return Address of first node in
// the loop if present else return NULL.
import java.io.*;
import java.util.*;

class GFG
{
  static Node detectCycle(Node A)
  {

    // declaring map to store node address
    Set<Node> uset = new HashSet<Node>();
    Node = A;

    // Default consider that no cycle is present
    while (ptr != NULL)
    {

      // checking if address is already present in map
      if(uset.contains(ptr))
      {
        return ptr;
      }

      // if address not present then insert into the set
      else
      {
        uset.add(ptr);
      }
      ptr = ptr.next;
    }
    return null;
  }
}

// This code is contributed by avanitrachhadiya2155

Python3

# The below function take head of Linked List as
# input and return Address of first node in
# the loop if present else return NULL.

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

    # Declaring map to store node
    # address
    uset = set()

    ptr = A

    # Default consider that no cycle
    # is present
    while (ptr != None):

        # Checking if address is already
        # present in map
        if (ptr in uset):
          return ptr

        # If address not present then
        # insert into the set
        else:
            uset.add(ptr)

        ptr = ptr.next

    return None

# This code is contributed by pratham76

C#

// The below function take head of Linked List as
// input and return Address of first node in
// the loop if present else return NULL.

using System.Collections.Generic;

public class GFG
{
class Node
{
    public int key;
    public Node next;
};
  static Node detectCycle(Node A)
  {

    // declaring map to store node address
    HashSet<Node> uset = new HashSet<Node>();
    Node ptr = A;

    // Default consider that no cycle is present
    while (ptr != null)
    {

      // checking if address is already present in map
      if(uset.Contains(ptr))
      {
        return ptr;
      }

      // if address not present then insert into the set
      else
      {
        uset.Add(ptr);
      }
      ptr = ptr.next;
    }
    return null;
  }
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// The below function take head of Linked List as
// input and return Address of first node in
// the loop if present else return NULL.

    function detectCycle(A)
    {
        // declaring map to store node address
        let uset = new Set();

        let ptr = A;
        // Default consider that no cycle is present
    while (ptr != NULL)
    {

      // checking if address is already present in map
      if(uset.has(ptr))
      {
        return ptr;
      }

      // if address not present then insert into the set
      else
      {
        uset.add(ptr);
      }
      ptr = ptr.next;
    }
    return null;
    }


// This code is contributed by patel2127

</script>

Output

Loop starting node is 15

Time Complexity: O(NlogN)
Auxiliary Space: O(N)