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.
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 stop 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 circularimport java.util.*;class GFG{ /* Link list Node */static class Node{ int data; Node next;}/*This function returns true if given linkedlist 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 stop 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 classclass 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 objectclass LinkedList: # Function to initialize head def __init__(self): self.head = Nonedef 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 hereif __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 circularusing System;public class GFG{ /* Link list Node */public class Node{ public int data; public Node next;}/*This function returns true if given linkedlist 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 stop 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 |
Javascript
<script>// javascript program to check if// linked list is circular /* Link list Node */class Node { constructor(val) { this.data = val; this.next = null; }} /* * This function returns true if given linked list is circular, else false. */ function isCircular( head) { // An empty linked list is circular if (head == null) return true; // Next of head node = head.next; // This loop would stop 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. function newNode(data) { temp = new Node(); temp.data = data; temp.next = null; return temp; } /* Driver code */ /* Start with the empty list */ head = newNode(1); head.next = newNode(2); head.next.next = newNode(3); head.next.next.next = newNode(4); document.write(isCircular(head) ? "Yes<br/>" : "No<br/>"); // Making linked list circular head.next.next.next.next = head; document.write(isCircular(head) ? "Yes<br/>" : "No<br/>");// This code contributed by gauravrajput1</script> |
Output :
No Yes
?list=PLqM7alHXFySH41ZxzrPNj2pAYPOI8ITe7
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