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# Detect loop in a linked list

• Difficulty Level : Easy
• Last Updated : 28 Jan, 2023

Given a linked list, check if the linked list has a loop or not. The below diagram shows a linked list with a loop.

## Detect loop in a linked list using Hashing:

The idea is to insert the nodes in the hashmap and whenever a node is encountered that is already present in the hashmap then return true.

Follow the steps below to solve the problem:

• Traverse the list individually and keep putting the node addresses in a Hash Table.
• At any point, if NULL is reached then return false
• If the next of the current nodes points to any of the previously stored nodes in  Hash then return true.

Below is the implementation of the above approach:

## C++

 // C++ program to detect loop in a linked list #include using namespace std;   /* Link list node */ struct Node {     int data;     struct Node* next; };   void push(struct Node** head_ref, int new_data) {     /* allocate node */     struct Node* new_node = new Node;       /* put in the data  */     new_node->data = new_data;       /* link the old list of the new node */     new_node->next = (*head_ref);       /* move the head to point to the new node */     (*head_ref) = new_node; }   // Returns true if there is a loop in linked list // else returns false. bool detectLoop(struct Node* h) {     unordered_set s;     while (h != NULL) {         // If this node is already present         // in hashmap it means there is a cycle         // (Because you will be encountering the         // node for the second time).         if (s.find(h) != s.end())             return true;           // If we are seeing the node for         // the first time, insert it in hash         s.insert(h);           h = h->next;     }       return false; }   /* Driver program to test above function*/ int main() {     /* Start with the empty list */     struct Node* head = NULL;       push(&head, 20);     push(&head, 4);     push(&head, 15);     push(&head, 10);       /* Create a loop for testing */     head->next->next->next->next = head;       if (detectLoop(head))         cout << "Loop Found";     else         cout << "No Loop";       return 0; } // This code is contributed by Geetanjali

## Java

 // Java program to detect loop in a linked list import java.util.*;   public class LinkedList {       static Node head; // head of list       /* Linked list Node*/     static class Node {         int data;         Node next;         Node(int d)         {             data = d;             next = null;         }     }       /* Inserts a new Node at front of the list. */     static public void push(int new_data)     {         /* 1 & 2: Allocate the Node &                   Put in the data*/         Node new_node = new Node(new_data);           /* 3. Make next of new Node as head */         new_node.next = head;           /* 4. Move the head to point to new Node */         head = new_node;     }       // Returns true if there is a loop in linked     // list else returns false.     static boolean detectLoop(Node h)     {         HashSet s = new HashSet();         while (h != null) {             // If we have already has this node             // in hashmap it means their is a cycle             // (Because you we encountering the             // node second time).             if (s.contains(h))                 return true;               // If we are seeing the node for             // the first time, insert it in hash             s.add(h);               h = h.next;         }           return false;     }       /* Driver program to test above function */     public static void main(String[] args)     {         LinkedList llist = new LinkedList();           llist.push(20);         llist.push(4);         llist.push(15);         llist.push(10);           /*Create loop for testing */         llist.head.next.next.next.next = llist.head;           if (detectLoop(head))             System.out.println("Loop Found");         else             System.out.println("No Loop");     } }   // This code is contributed by Arnav Kr. Mandal.

## Python3

 # Python3 program to detect loop # in the linked list   # Node class class Node:       # Constructor to initialize     # the node object     def __init__(self, data):         self.data = data         self.next = None     class LinkedList:       # Function to initialize head     def __init__(self):         self.head = None       # Function to insert a new     # node at the beginning     def push(self, new_data):         new_node = Node(new_data)         new_node.next = self.head         self.head = new_node       # Utility function to print it     # the linked LinkedList     def printList(self):         temp = self.head         while(temp):             print(temp.data, end=" ")             temp = temp.next       def detectLoop(self):         s = set()         temp = self.head         while (temp):               # If we already have             # this node in hashmap it             # means there is a cycle             # (Because we are encountering             # the node second time).             if (temp in s):                 return True               # If we are seeing the node for             # the first time, insert it in hash             s.add(temp)               temp = temp.next           return False     # Driver program for testing llist = LinkedList() llist.push(20) llist.push(4) llist.push(15) llist.push(10)   # Create a loop for testing llist.head.next.next.next.next = llist.head   if(llist.detectLoop()):     print("Loop Found") else:     print("No Loop ")   # This code is contributed by Gitanjali.

## C#

 // C# program to detect loop in a linked list using System; using System.Collections.Generic;   class LinkedList {       // head of list     public Node head;       /* Linked list Node*/     public class Node {         public int data;         public Node next;         public Node(int d)         {             data = d;             next = null;         }     }       /* Inserts a new Node at front of the list. */     public void push(int new_data)     {         /* 1 & 2: Allocate the Node &                 Put in the data*/         Node new_node = new Node(new_data);           /* 3. Make next of new Node as head */         new_node.next = head;           /* 4. Move the head to point to new Node */         head = new_node;     }       // Returns true if there is a loop in linked     // list else returns false.     public static bool detectLoop(Node h)     {         HashSet s = new HashSet();         while (h != null) {             // If we have already has this node             // in hashmap it means their is a cycle             // (Because you we encountering the             // node second time).             if (s.Contains(h))                 return true;               // If we are seeing the node for             // the first time, insert it in hash             s.Add(h);               h = h.next;         }           return false;     }       /* Driver code*/     public static void Main(String[] args)     {         LinkedList llist = new LinkedList();           llist.push(20);         llist.push(4);         llist.push(15);         llist.push(10);           /*Create loop for testing */         llist.head.next.next.next.next = llist.head;           if (detectLoop(llist.head))             Console.WriteLine("Loop Found");         else             Console.WriteLine("No Loop");     } }   // This code has been contributed by 29AjayKumar

## Javascript



Output

Loop Found

Time complexity: O(N), Only one traversal of the loop is needed.
Auxiliary Space: O(N), N is the space required to store the value in the hashmap.

## Detect loop in a linked list by Modification In Node Structure:

The idea is to modify the node structure by adding flag in it and mark the flag whenever visit the node.

Follow the steps below to solve the problem:

• Have a visited flag with each node.
• Traverse the linked list and keep marking visited nodes.
• If you see a visited node again then there is a loop.

Below is the implementation of the above approach.

## C++

 // C++ program to detect loop in a linked list #include using namespace std;   /* Link list node */ struct Node {     int data;     struct Node* next;     int flag; };   void push(struct Node** head_ref, int new_data) {     /* allocate node */     struct Node* new_node = new Node;       /* put in the data */     new_node->data = new_data;       new_node->flag = 0;       /* link the old list of the new node */     new_node->next = (*head_ref);       /* move the head to point to the new node */     (*head_ref) = new_node; }   // Returns true if there is a loop in linked list // else returns false. bool detectLoop(struct Node* h) {     while (h != NULL) {         // If this node is already traverse         // it means there is a cycle         // (Because you we encountering the         // node for the second time).         if (h->flag == 1)             return true;           // If we are seeing the node for         // the first time, mark its flag as 1         h->flag = 1;           h = h->next;     }       return false; }   /* Driver program to test above function*/ int main() {     /* Start with the empty list */     struct Node* head = NULL;       push(&head, 20);     push(&head, 4);     push(&head, 15);     push(&head, 10);       /* Create a loop for testing */     head->next->next->next->next = head;       if (detectLoop(head))         cout << "Loop Found";     else         cout << "No Loop";       return 0; } // This code is contributed by Geetanjali

## Javascript



Output

Loop Found

Time complexity: O(N), Only one traversal of the loop is needed.
Auxiliary Space: O(1)

## Detect loop in a linked list using Floyd’s Cycle-Finding Algorithm:

This algorithm is used to find a loop in a linked list. It uses two pointers one moving twice as fast as the other one. The faster one is called the faster pointer and the other one is called the slow pointer.

Follow the steps below to solve the problem:

• Traverse linked list using two pointers.
• Move one pointer(slow_p) by one and another pointer(fast_p) by two.
• If these pointers meet at the same node then there is a loop. If pointers do not meet then the linked list doesn’t have a loop.

Illustration:

The below image shows how the detect loop function works in the code:

Implementation of Floyd’s Cycle-Finding Algorithm:

## Python

 // C++ program to detect loop in a linked list #include using namespace std;   /* Link list node */ class Node { public:     int data;     Node* next; };   void push(Node** head_ref, int new_data) {     /* allocate node */     Node* new_node = new Node();       /* put in the data */     new_node->data = new_data;       /* link the old list of the new node */     new_node->next = (*head_ref);       /* move the head to point to the new node */     (*head_ref) = new_node; }   int detectLoop(Node* list) {     Node *slow_p = list, *fast_p = list;       while (slow_p && fast_p && fast_p->next) {         slow_p = slow_p->next;         fast_p = fast_p->next->next;         if (slow_p == fast_p) {             return 1;         }     }     return 0; }   /* Driver code*/ int main() {     /* Start with the empty list */     Node* head = NULL;       push(&head, 20);     push(&head, 4);     push(&head, 15);     push(&head, 10);       /* Create a loop for testing */     head->next->next->next->next = head;     if (detectLoop(head))         cout << "Loop Found";     else         cout << "No Loop";     return 0; }   // This is code is contributed by rathbhupendra

## C

 // C program to detect loop in a linked list #include #include   /* Link list node */ struct Node {     int data;     struct Node* next; };   void push(struct Node** head_ref, int new_data) {     /* allocate node */     struct Node* new_node         = (struct Node*)malloc(sizeof(struct Node));       /* put in the data  */     new_node->data = new_data;       /* link the old list of the new node */     new_node->next = (*head_ref);       /* move the head to point to the new node */     (*head_ref) = new_node; }   int detectLoop(struct Node* list) {     struct Node *slow_p = list, *fast_p = list;       while (slow_p && fast_p && fast_p->next) {         slow_p = slow_p->next;         fast_p = fast_p->next->next;         if (slow_p == fast_p) {             return 1;         }     }     return 0; }   /* Driver program to test above function*/ int main() {     /* Start with the empty list */     struct Node* head = NULL;       push(&head, 20);     push(&head, 4);     push(&head, 15);     push(&head, 10);       /* Create a loop for testing */     head->next->next->next->next = head;       if (detectLoop(head))         printf("Loop Found");     else         printf("No Loop");     return 0; }

## Java

 // Java program to detect loop in a linked list   import java.io.*;   class LinkedList {     Node head; // head of list       /* Linked list Node*/     class Node {         int data;         Node next;         Node(int d)         {             data = d;             next = null;         }     }       /* Inserts a new Node at front of the list. */     public void push(int new_data)     {         /* 1 & 2: Allocate the Node &                 Put in the data*/         Node new_node = new Node(new_data);           /* 3. Make next of new Node as head */         new_node.next = head;           /* 4. Move the head to point to new Node */         head = new_node;     }       void detectLoop()     {         Node slow_p = head, fast_p = head;         int flag = 0;         while (slow_p != null && fast_p != null                && fast_p.next != null) {             slow_p = slow_p.next;             fast_p = fast_p.next.next;             if (slow_p == fast_p) {                 flag = 1;                 break;             }         }         if (flag == 1)             System.out.println("Loop Found");         else             System.out.println("No Loop");     }       /* Driver program to test above functions */     public static void main(String args[])     {         LinkedList llist = new LinkedList();           llist.push(20);         llist.push(4);         llist.push(15);         llist.push(10);           /*Create loop for testing */         llist.head.next.next.next.next = llist.head;           llist.detectLoop();     } } /* This code is contributed by Rajat Mishra. */

## Python

 # Python program to detect loop in the linked list   # Node class     class Node:       # Constructor to initialize the node object     def __init__(self, data):         self.data = data         self.next = None     class LinkedList:       # Function to initialize head     def __init__(self):         self.head = None       # Function to insert a new node at the beginning     def push(self, new_data):         new_node = Node(new_data)         new_node.next = self.head         self.head = new_node       # Utility function to print it the linked LinkedList     def printList(self):         temp = self.head         while(temp):             print(temp.data)             temp = temp.next       def detectLoop(self):         slow_p = self.head         fast_p = self.head         while(slow_p and fast_p and fast_p.next):             slow_p = slow_p.next             fast_p = fast_p.next.next             if slow_p == fast_p:                 return 1         return 0     # Driver program for testing llist = LinkedList() llist.push(20) llist.push(4) llist.push(15) llist.push(10)   # Create a loop for testing llist.head.next.next.next.next = llist.head if(llist.detectLoop()):     print("Loop Found") else:     print("No Loop")   # This code is contributed by Nikhil Kumar Singh(nickzuck_007)

## C#

 // C# program to detect loop in a linked list using System;   public class LinkedList {     Node head; // head of list       /* Linked list Node*/     public class Node {         public int data;         public Node next;         public Node(int d)         {             data = d;             next = null;         }     }       /* Inserts a new Node at front of the list. */     public void push(int new_data)     {         /* 1 & 2: Allocate the Node &                 Put in the data*/         Node new_node = new Node(new_data);           /* 3. Make next of new Node as head */         new_node.next = head;           /* 4. Move the head to point to new Node */         head = new_node;     }       Boolean detectLoop()     {         Node slow_p = head, fast_p = head;         while (slow_p != null && fast_p != null                && fast_p.next != null) {             slow_p = slow_p.next;             fast_p = fast_p.next.next;             if (slow_p == fast_p) {                 return true;             }         }         return false;     }       /* Driver code */     public static void Main(String[] args)     {         LinkedList llist = new LinkedList();           llist.push(20);         llist.push(4);         llist.push(15);         llist.push(10);         /*Create loop for testing */         llist.head.next.next.next.next = llist.head;           Boolean found = llist.detectLoop();         if (found) {             Console.WriteLine("Loop Found");         }         else {             Console.WriteLine("No Loop");         }     } }   // This code is contributed by Princi Singh

## Javascript



Output

Loop Found

Time complexity: O(N), Only one traversal of the loop is needed.
Auxiliary Space: O(1).

## Detect loop in a linked list by Marking visited nodes without modifying Node structure:

The idea is to point the current node of the linked list to a node which is created. Whenever a node’s next is pointing to that node it means loop is there.

Follow the steps below to solve the problem:

• 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 find that loop exists.
• If we come across a node that points to null then the loop doesn’t exist.

Below is the implementation of the above approach:

## Javascript



Output

Loop Found

Time complexity: O(N). Only one traversal of the loop is needed.
Auxiliary Space: O(1)

## Detect loop in a linked list by Storing length:

The idea is to store the length of the list from first node and last node, increment last node till reaches NULL or number of nodes in last is greater the current between first and last nodes.

Follow the steps below to solve the problem:

• In this method, two pointers are created, first (always points to head) and last.
• Each time the last pointer moves calculate no of nodes between the first and last.
• check whether the current no of nodes > previous no of nodes
• If yes we proceed by moving the last pointer
• Else it means we’ve reached the end of the loop, so return output accordingly.

Below is the implementation of the above approach:

## C++

 // C++ program to return first node of loop #include 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; }   /*returns distance between first and last node every time  * last node moves forwards*/ int distance(Node* first, Node* last) {     /*counts no of nodes between first and last*/     int counter = 0;       Node* curr;     curr = first;       while (curr != last) {         counter += 1;         curr = curr->next;     }       return counter + 1; }   // Function to detect first node of loop // in a linked list that may contain loop bool detectLoop(Node* head) {       // Create a temporary node     Node* temp = new Node;       Node *first, *last;       /*first always points to head*/     first = head;     /*last pointer initially points to head*/     last = head;       /*current_length stores no of nodes between current      * position of first and last*/     int current_length = 0;       /*prev_length stores no of nodes between previous      * position of first and last*/     int prev_length = -1;       while (current_length > prev_length && last != NULL) {         // set prev_length to current length then update the         // current length         prev_length = current_length;         // distance is calculated         current_length = distance(first, last);         // last node points the next node         last = last->next;     }       if (last == NULL) {         return false;     }     else {         return true;     } }   /* Driver program to test above function*/ int main() {     Node* head = newNode(1);     head->next = newNode(2);     head->next->next = newNode(3);     head->next->next->next = newNode(4);     head->next->next->next->next = newNode(5);       /* Create a loop for testing(5 is pointing to 3) */     head->next->next->next->next->next = head->next->next;       bool found = detectLoop(head);     if (found)         cout << "Loop Found";     else         cout << "No Loop Found";       return 0; }

## 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(); }   /*returns distance between first and last node every time  * last node moves forwards*/ static int distance(Node first, Node last) {         /*counts no of nodes between first and last*/     int counter = 0;     Node curr;     curr = first;     while (curr != last)     {         counter += 1;         curr = curr.next;     }     return counter + 1; }   // Function to detect first node of loop // in a linked list that may contain loop static boolean detectLoop(Node head) {       // Create a temporary node     Node temp = new Node();     Node first, last;       /*first always points to head*/     first = head;         /*last pointer initially points to head*/     last = head;       /*current_length stores no of nodes between current      * position of first and last*/     int current_length = 0;       /*current_length stores no of nodes between previous      * position of first and last*/     int prev_length = -1;     while (current_length > prev_length && last != null)     {                 // set prev_length to current length then update the         // current length           prev_length = current_length;                 // distance is calculated         current_length = distance(first, last);                 // last node points the next node         last = last.next;     }           if (last == null)     {         return false;     }     else     {         return true;     } }   /* Driver program to test above function*/ public static void main(String[] args) {     Node head = newNode(1);     head.next = newNode(2);     head.next.next = newNode(3);     head.next.next.next = newNode(4);     head.next.next.next.next = newNode(5);       /* Create a loop for testing(5 is pointing to 3) */     head.next.next.next.next.next = head.next.next;     boolean found = detectLoop(head);     if (found)         System.out.print("Loop Found");     else         System.out.print("No Loop Found"); } }   // This code is contributed by gauravrajput1

## Python3

 # Python program to return first node of loop class newNode:     def __init__(self, key):         self.key = key         self.left = None         self.right = None             # A utility function to print a linked list def printList(head):       while (head != None) :         print(head.key, end=" ")         head = head.next;           print()     # returns distance between first and last node every time # last node moves forwards def distance(first, last):       # counts no of nodes between first and last     counter = 0       curr = first       while (curr != last):         counter = counter + 1         curr = curr.next             return counter + 1       # Function to detect first node of loop # in a linked list that may contain loop def detectLoop(head):       # Create a temporary node     temp = ""       # first always points to head     first = head;     # last pointer initially points to head     last = head;       # current_length stores no of nodes between current     # position of first and last     current_length = 0       #current_length stores no of nodes between previous     # position of first and last*/     prev_length = -1       while (current_length > prev_length and last != None) :         # set prev_length to current length then update the         # current length         prev_length = current_length         # distance is calculated         current_length = distance(first, last)         # last node points the next node         last = last.next;                 if (last == None) :         return False           else :         return True      # Driver program to test above function   head = newNode(1); head.next = newNode(2); head.next.next = newNode(3); head.next.next.next = newNode(4); head.next.next.next.next = newNode(5);   # Create a loop for testing(5 is pointing to 3) head.next.next.next.next.next = head.next.next;   found = detectLoop(head) if (found) :     print("Loop Found") else :     print("No Loop Found")   # This code is contributed by ihritik

## C#

 // C# program to return first node of loop using System;   public class GFG {   public     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();   }     /*returns distance between first and last node every time  * last node moves forwards*/   static int distance(Node first, Node last)   {       /*counts no of nodes between first and last*/     int counter = 0;     Node curr;     curr = first;     while (curr != last)     {       counter += 1;       curr = curr.next;     }     return counter + 1;   }     // Function to detect first node of loop   // in a linked list that may contain loop   static bool detectLoop(Node head)   {       // Create a temporary node     Node temp = new Node();     Node first, last;       /*first always points to head*/     first = head;       /*last pointer initially points to head*/     last = head;       /*current_length stores no of nodes between current      * position of first and last*/     int current_length = 0;       /*current_length stores no of nodes between previous      * position of first and last*/     int prev_length = -1;     while (current_length > prev_length && last != null)     {         // set prev_length to current length then update the       // current length       prev_length = current_length;         // distance is calculated       current_length = distance(first, last);         // last node points the next node       last = last.next;     }       if (last == null)     {       return false;     }     else     {       return true;     }   }     /* Driver program to test above function*/   public static void Main(String[] args)   {     Node head = newNode(1);     head.next = newNode(2);     head.next.next = newNode(3);     head.next.next.next = newNode(4);     head.next.next.next.next = newNode(5);       /* Create a loop for testing(5 is pointing to 3) */     head.next.next.next.next.next = head.next.next;     bool found = detectLoop(head);     if (found)       Console.Write("Loop Found");     else       Console.Write("No Loop Found");   } }   // This code is contributed by gauravrajput1

## Javascript



Output

Loop Found

Time complexity: O(N2), For every node calculate the length of that node from the head by traversing.
Auxiliary Space: O(1)

## Detect loop in a linked list by Modifying Value:

The idea is to modify the value of the visited node and check if current nodes value is equal to that value or not.

Follow the steps below to solve the problem:

• Traverse the linked list and change the value(data) of that node to -1.
• Now, before modifying the value check whether the value of the node is already -1
• If it is -1 then return TRUE
• Otherwise, change the value of the node.
• Traverse the linked list till it reaches NULL.

Below is the implementation of the above approach:

## Javascript



Output

Loop Found

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