Check if a linked list is Circular Linked List

Given a singly linked list, find if the linked list is circular or not. A linked list is called circular if it is not NULL-terminated and all nodes are connected in the form of a cycle. Below is an example of a circular linked list.

An empty linked list is considered as circular.

Note that this problem is different from cycle detection problem, here all nodes have to be part of cycle.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

The idea is to store head of the linked list and traverse it. If we reach NULL, linked list is not circular. If reach head again, linked list is circular.

C++

// C++ program to check if linked list is circular 
#include<bits/stdc++.h> 
using namespace std; 
  
/* Link list Node */
struct Node 
{ 
    int data; 
    struct Node* next; 
}; 
  
/* This function returns true if given linked 
   list is circular, else false. */
bool isCircular(struct Node *head) 
{ 
    // An empty linked list is circular 
    if (head == NULL) 
       return true; 
  
    // Next of head 
    struct Node *node = head->next; 
  
    // This loop would stope in both cases (1) If 
    // Circular (2) Not circular 
    while (node != NULL && node != head) 
       node = node->next; 
  
    // If loop stopped because of circular 
    // condition 
    return (node == head); 
} 
  
// Utility function to create a new node. 
Node *newNode(int data) 
{ 
    struct Node *temp = new Node; 
    temp->data = data; 
    temp->next = NULL; 
    return temp; 
} 
  
/* Driver program to test above function*/
int main() 
{ 
    /* Start with the empty list */
    struct Node* head = newNode(1); 
    head->next = newNode(2); 
    head->next->next = newNode(3); 
    head->next->next->next = newNode(4); 
  
    isCircular(head)? cout << "Yes\n" : 
                      cout << "No\n" ; 
  
    // Making linked list circular 
    head->next->next->next->next = head; 
  
    isCircular(head)? cout << "Yes\n" : 
                      cout << "No\n" ; 
  
    return 0; 
} 

Java

// Java program to check if  
// linked list is circular 
import java.util.*; 
  
class GFG 
{ 
      
/* Link list Node */
static class Node 
{ 
    int data; 
    Node next; 
} 
  
/*This function returns true if given linked 
list is circular, else false. */
static boolean isCircular( Node head) 
{ 
    // An empty linked list is circular 
    if (head == null) 
    return true; 
  
    // Next of head 
    Node node = head.next; 
  
    // This loop would stope in both cases (1) If 
    // Circular (2) Not circular 
    while (node != null && node != head) 
    node = node.next; 
  
    // If loop stopped because of circular 
    // condition 
    return (node == head); 
} 
  
// Utility function to create a new node. 
static Node newNode(int data) 
{ 
    Node temp = new Node(); 
    temp.data = data; 
    temp.next = null; 
    return temp; 
} 
  
/* Driver code*/
public static void main(String args[]) 
{ 
    /* Start with the empty list */
    Node head = newNode(1); 
    head.next = newNode(2); 
    head.next.next = newNode(3); 
    head.next.next.next = newNode(4); 
  
    System.out.print(isCircular(head)? "Yes\n" : 
                    "No\n" ); 
  
    // Making linked list circular 
    head.next.next.next.next = head; 
  
    System.out.print(isCircular(head)? "Yes\n" : 
                    "No\n" ); 
  
} 
} 
  
// This code contributed by Arnab Kundu 

Python3

# A simple Python program to check if a linked list is circular 
  
# Node class  
class Node:  
  
    # Function to initialise the node object  
    def __init__(self, data):  
        self.data = data # Assign data  
        self.next = None # Initialize next as null  
  
  
# Linked List class contains a Node object  
class LinkedList:  
  
    # Function to initialize head  
    def __init__(self):  
        self.head = None
  
def Circular(head): 
    if head==None: 
        return True
          
    # Next of head 
    node = head.next
    i = 0
      
    # This loop would stop in both cases (1) If 
    # Circular (2) Not circular 
    while((node is not None) and (node is not head)): 
        i = i + 1
        node = node.next
      
    return(node==head) 
  
  
# Code execution starts here  
if __name__=='__main__': 
    llist = LinkedList()  
    llist.head = Node(1) 
    second = Node(2) 
    third = Node(3)  
    fourth = Node(4) 
      
    llist.head.next = second; 
    second.next = third; 
    third.next = fourth 
      
    if (Circular(llist.head)): 
        print('Yes') 
    else: 
        print('No') 
      
    fourth.next = llist.head 
      
    if (Circular(llist.head)): 
        print('Yes') 
    else: 
        print('No') 
          
# This code is contributed by Sanket Badhe 

C#

// C# program to check if  
// linked list is circular  
using System; 
public class GFG  
{  
      
/* Link list Node */
public class Node  
{  
    public int data;  
    public Node next;  
}  
  
/*This function returns true if given linked  
list is circular, else false. */
static bool isCircular( Node head)  
{  
    // An empty linked list is circular  
    if (head == null)  
    return true;  
  
    // Next of head  
    Node node = head.next;  
  
    // This loop would stope in both cases (1) If  
    // Circular (2) Not circular  
    while (node != null && node != head)  
    node = node.next;  
  
    // If loop stopped because of circular  
    // condition  
    return (node == head);  
}  
  
// Utility function to create a new node.  
static Node newNode(int data)  
{  
    Node temp = new Node();  
    temp.data = data;  
    temp.next = null;  
    return temp;  
}  
  
/* Driver code*/
public static void Main(String []args)  
{  
    /* Start with the empty list */
    Node head = newNode(1);  
    head.next = newNode(2);  
    head.next.next = newNode(3);  
    head.next.next.next = newNode(4);  
  
    Console.Write(isCircular(head)? "Yes\n" :  
                    "No\n" );  
  
    // Making linked list circular  
    head.next.next.next.next = head;  
  
    Console.Write(isCircular(head)? "Yes\n" :  
                    "No\n" );  
  
}  
}  
// This code has been contributed by 29AjayKumar 

Output :

No
Yes

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My Personal Notes

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python3

C#

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