Given a linked list, check if the linked list has loop or not. Below diagram shows a linked list with a loop.
Following are different ways of doing this
Solution 1: Hashing Approach:
Traverse the list one by one and keep putting the node addresses in a Hash Table. At any point, if NULL is reached then return false and if next of current node points to any of the previously stored nodes in Hash then return true.
C++
// C++ program to detect loop in a linked list #include <bits/stdc++.h> 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 off 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<Node*> s; while (h != NULL) { // If this node is already present // in hashmap it means there is a cycle // (Because you we 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<Node> s = new HashSet<Node>(); 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 have already has # this node in hashmap it # means their is a cycle # (Because you we 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<Node> s = new HashSet<Node>(); 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 |
Loop found
Complexity Analysis:
- 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 hashmap.
Solution 2: This problem can be solved without hashmap by modifying the linked list data-structure.
Approach: This solution requires modifications to the basic linked list data structure.
- 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. This solution works in O(n) but requires additional information with each node.
- A variation of this solution that doesn’t require modification to basic data structure can be implemented using a hash, just store the addresses of visited nodes in a hash and if you see an address that already exists in hash then there is a loop.
C++
// C++ program to detect loop in a linked list #include <bits/stdc++.h> 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 off 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 |
Java
// Java program to detect loop in a linked list import java.util.*; class GFG{ // Link list node static class Node { int data; Node next; int flag; }; static Node push(Node head_ref, int new_data) { // Allocate node Node new_node = new Node(); // Put in the data new_node.data = new_data; new_node.flag = 0 ; // Link the old list off the new node new_node.next = head_ref; // Move the head to point to the new node head_ref = new_node; return head_ref; } // Returns true if there is a loop in linked // list else returns false. static boolean detectLoop(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 code public static void main(String[] args) { // Start with the empty list Node head = null ; head = push(head, 20 ); head = push(head, 4 ); head = push(head, 15 ); head = push(head, 10 ); // Create a loop for testing head.next.next.next.next = head; if (detectLoop(head)) System.out.print( "Loop found" ); else System.out.print( "No Loop" ); } } // This code is contributed by Rajput-Ji |
Python3
# Python3 program to detect loop in a linked list ''' Link list node ''' class Node: def __init__( self ): self .data = 0 self . next = None self .flag = 0 def push(head_ref, new_data): ''' allocate node ''' new_node = Node(); ''' put in the data ''' new_node.data = new_data; new_node.flag = 0 ; ''' link the old list off the new node ''' new_node. next = (head_ref); ''' move the head to point to the new node ''' (head_ref) = new_node; return head_ref # Returns true if there is a loop in linked list # else returns false. def detectLoop(h): while (h ! = None ): # 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''' if __name__ = = '__main__' : ''' Start with the empty list ''' head = None ; head = push(head, 20 ); head = push(head, 4 ); head = push(head, 15 ); head = push( head, 10 ) ''' Create a loop for testing ''' head. next . next . next . next = head; if (detectLoop(head)): print ( "Loop found" ) else : print ( "No Loop" ) # This code is contributed by rutvik_56 |
C#
// C# program to detect loop in a linked list using System; class GFG{ // Link list node class Node { public int data; public Node next; public int flag; }; static Node push(Node head_ref, int new_data) { // Allocate node Node new_node = new Node(); // Put in the data new_node.data = new_data; new_node.flag = 0; // Link the old list off the new node new_node.next = head_ref; // Move the head to point to the new node head_ref = new_node; return head_ref; } // Returns true if there is a loop in linked // list else returns false. static bool detectLoop(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 code public static void Main( string [] args) { // Start with the empty list Node head = null ; head = push(head, 20); head = push(head, 4); head = push(head, 15); head = push(head, 10); // Create a loop for testing head.next.next.next.next = head; if (detectLoop(head)) Console.Write( "Loop found" ); else Console.Write( "No Loop" ); } } // This code is contributed by pratham76 |
Loop Found
Complexity Analysis:
- Time complexity:O(n).
Only one traversal of the loop is needed. - Auxiliary Space:O(1).
No extra space is needed.
Solution 3: Floyd’s Cycle-Finding Algorithm
Approach: This is the fastest method and has been described below:
- 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 linked list doesn’t have a loop.
Below image shows how the detectloop function works in the code :
Implementation of Floyd’s Cycle-Finding Algorithm:
C++
// C++ program to detect loop in a linked list #include <bits/stdc++.h> 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 off 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 <stdio.h> #include <stdlib.h> /* 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 off 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 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( "Loop not found" ); } /* 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 # 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 "Found Loop" 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 |
Found Loop
Complexity Analysis:
- Time complexity: O(n).
Only one traversal of the loop is needed. - Auxiliary Space:O(1).
There is no space required.
How does above algorithm work?
Please See : How does Floyd’s slow and fast pointers approach work?
https://www.youtube.com/watch?v=Aup0kOWoMVg
References:
http://en.wikipedia.org/wiki/Cycle_detection
http://ostermiller.org/find_loop_singly_linked_list.html
Solution 4: Marking visited nodes without modifying the linked list data structure
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 find that loop exists. If we come across a node that points to null then loop doesn’t exist.
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 bool 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 false ; } // Check if next is already // pointing to temp if (head->next == temp) { return true ; } // 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 false ; } /* 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" ; return 0; } |
Java
// Java program to return first node of loop 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 boolean 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 false ; } // Check if next is already // pointing to temp if (head.next == temp) { return true ; } // 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 false ; } // Driver code 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.println( "Loop Found" ); else System.out.println( "No Loop" ); } } // This code is contributed by Arnab Kundu |
Python3
# Python3 program to return first node of loop # A binary tree node has data, pointer to # left child and a pointer to right child # Helper function that allocates a new node # with the given data and None left and # right pointers class newNode: def __init__( self , key): self .key = key self .left = None self .right = None # A utility function to pra linked list def printList(head): while (head ! = None ): print (head.key, end = " " ) head = head. next print () # Function to detect first node of loop # in a linked list that may contain loop def detectLoop(head): # Create a temporary node temp = "" while (head ! = None ): # This condition is for the case # when there is no loop if (head. next = = None ): return False # Check if next is already # pointing to temp if (head. next = = temp): return True # Store the pointer to the next node # in order to get to it in the next step nex = head. next # Make next poto temp head. next = temp # Get to the next node in the list head = nex return False # Driver Code 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" ) # This code is contributed by SHUBHAMSINGH10 |
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(); } // 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(); while (head != null ) { // This condition is for the case // when there is no loop if (head.next == null ) { return false ; } // Check if next is already // pointing to temp if (head.next == temp) { return true ; } // 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 false ; } // Driver code 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) { Console.WriteLine( "Loop Found" ); } else { Console.WriteLine( "No Loop" ); } } } // This code is contributed by Princi Singh |
Loop Found
Complexity Analysis:
- Time complexity: O(n).
Only one traversal of the loop is needed. - Auxiliary Space: O(1).
There is no space required.
Solution 5: Store length
In this method, two pointers are created, first (always points to head) and last. Each time last pointer moves we calculate no of nodes in between first and last and check whether current no of nodes > previous no of nodes, if yes we proceed by moving last pointer else it means we’ve reached the end of the loop, so we return output accordingly.
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; } /*returns distance between first and last node every time * last node moves forwars*/ 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; /*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*/ 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; } |
Loop Found
Complexity Analysis:
- Time complexity: O(n2)
- Auxiliary Space: O(1)
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.