Detect loop in a linked list
Given a linked list, check if the linked list has loop or not. Below diagram shows a linked list with a loop.
The 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 the next of the current nodes 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 listimport 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 classclass Node: # Constructor to initialize # the node object def __init__(self, data): self.data = data self.next = Noneclass 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 testingllist = LinkedList()llist.push(20)llist.push(4)llist.push(15)llist.push(10)# Create a loop for testingllist.head.next.next.next.next = llist.headif(llist.detectLoop()): print("Loop found")else: print("No Loop ")# This code is contributed by Gitanjali. |
C#
// C# program to detect loop in a linked listusing 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 |
Javascript
<script>// JavaScript program to detect loop in a linked list var head; // head of list /* Linked list Node */ class Node { constructor(val) { this.data = val; this.next = null; } } /* Inserts a new Node at front of the list. */ function push(new_data) { /* * 1 & 2: Allocate the Node & Put in the data */var new_node = new Node(new_data); /* 3. Make next of new Node as head */ new_node.next = head; /* 4. Move the head to povar to new Node */ head = new_node; } // Returns true if there is a loop in linked // list else returns false. function detectLoop(h) { var s = new Set(); 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.has(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 */ push(20); push(4); push(15); push(10); /* Create loop for testing */ head.next.next.next.next = head; if (detectLoop(head)) document.write("Loop found"); else document.write("No Loop");// This code is contributed by todaysgaurav</script> |
Output
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 listimport java.util.*;class GFG{// Link list nodestatic 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 codepublic 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 listusing System;class GFG{// Link list nodeclass 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 codepublic 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 |
Javascript
<script>// JavaScript program to detect loop in a linked list// Link list nodeclass Node{ constructor() { let data; let next; let flag; }}function push( head_ref, new_data){ // Allocate node let 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. function detectLoop(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// Start with the empty list let 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)) document.write("Loop found"); else document.write("No Loop");// This code is contributed by rag2127</script> |
Output
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.
The 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 listclass 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 classclass Node: # Constructor to initialize the node object def __init__(self, data): self.data = data self.next = Noneclass 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 testingllist = LinkedList()llist.push(20)llist.push(4)llist.push(15)llist.push(10)# Create a loop for testingllist.head.next.next.next.next = llist.headif(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 listusing 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
<script>// Javascript program to detect loop in a linked listlet head; // head of list/* Linked list Node*/class Node{ constructor(d) { this.data = d; this.next = null; }}/* Inserts a new Node at front of the list. */function push(new_data){ /* 1 & 2: Allocate the Node & Put in the data*/ let 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;}function detectLoop(){ let slow_p = head, fast_p = head; let 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) document.write("Loop found<br>"); else document.write("Loop not found<br>");}// Driver codepush(20);push(4);push(15);push(10);// Create loop for testinghead.next.next.next.next = head;detectLoop();// This code is contributed by avanitrachhadiya2155</script> |
Output
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.
How does above algorithm work?
Please See : How does Floyd’s slow and fast pointers approach work?
https://www.youtube.com/watch?v=Aup0kOWoMVg
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 the 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 listvoid 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 loopbool 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 loopclass 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 pointersclass newNode: def __init__(self, key): self.key = key self.left = None self.right = None# A utility function to pra linked listdef 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 loopdef 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 Codehead = 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.nextfound = 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 loopusing 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 |
Javascript
<script>// Javascript program to return first node of loopclass Node{ constructor(key) { this.key = key; this.nexy = null; }}// A utility function to print a linked listfunction 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 loopfunction detectLoop(head){ // Create a temporary node let 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 let nex = head.next; // Make next point to temp head.next = temp; // Get to the next node in the list head = nex; } return false;}// Driver codelet head = new Node(1);head.next = new Node(2);head.next.next = new Node(3);head.next.next.next = new Node(4);head.next.next.next.next = new Node(5);// Create a loop for testing(5 is pointing to 3) /head.next.next.next.next.next = head.next.next;let found = detectLoop(head);if (found) document.write("Loop Found");else document.write("No Loop");// This code is contributed by ab2127</script> |
Output
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 the last pointer moves we calculate no of nodes in between first and last and check whether the 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 listvoid 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 loopbool 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;} |
Java
// Java program to return first node of loopimport 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 liststatic 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 forwars*/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 loopstatic 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 |
C#
// C# program to return first node of loopusing 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 forwars*/ 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
<script>// Javascript program to return first node of loopclass Node{ constructor(key) { this.key = key; this.next = null; }}function newNode(key){ let temp = new Node(key); return temp;}// A utility function to print a linked listfunction printList(head){ while (head != null) { document.write(head.key + " "); head = head.next; } document.write("</br>");}/*returns distance between first and lastnode every time last node moves forwars*/function distance(first, last){ /*counts no of nodes between first and last*/ let counter = 0; let 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 loopfunction detectLoop(head){ // Create a temporary node let temp = new Node(); let 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*/ let current_length = 0; /*current_length stores no of nodes between previous position of first and last*/ let 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 codelet 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;let found = detectLoop(head);if (found) document.write("Loop Found");else document.write("No Loop Found"); // This code is contributed by divyeshrabadiya07</script> |
Output
Loop Found
Complexity Analysis:
- Time complexity: O(n2)
- Auxiliary Space: O(1)
Another Approach:
- This is the simplest approach of the given problem, the only thing we have to do is to assign a new value to each data of node in the linked list which is not in the range given.
- Example suppose (1 <= Data on Node <= 10^3) then after visiting node assign the data as -1 as it is out of the given range.
Follow the code given below for a better understanding:
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;}// Function to detect first node of loop// in a linked list that may contain loopbool detectLoop(Node* head){ // If the head is null we will return false if (!head) return 0; else { // Traversing the linked list // for detecting loop while (head) { // If loop found if (head->key == -1) { return true; } // Changing the data of visited node to any // value which is outside th given range here it // is supposed the given range is (1 <= Data on // Node <= 10^3) else { head->key = -1; head = head->next; } } // If loop not found return false return 0; }}/* 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); cout << found << endl; return 0;} |
Java
// Java program to return first node of loopimport java.util.*;class LinkedList{ // Head of liststatic Node head;// Linked list Nodestatic 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;}// Function to detect first node of loop// in a linked list that may contain loopstatic boolean detectLoop(Node h){ // If the head is null we will return false if (head == null) return false; else { // Traversing the linked list // for detecting loop while (head != null) { // If loop found if (head.data == -1) { return true; } // Changing the data of visited node to any // value which is outside th given range // here it is supposed the given range is (1 // <= Data on Node <= 10^3) else { head.data = -1; head = head.next; } } // If loop not found return false return false; }}// Driver Codepublic static void main(String[] args){ LinkedList llist = new LinkedList(); llist.push(1); llist.push(2); llist.push(3); llist.push(4); llist.push(5); /* Create a loop for testing */ llist.head.next.next.next.next.next = llist.head.next.next; if (detectLoop(llist.head)) System.out.println("1"); else System.out.println("0");}}// This code is contributed by RohitOberoi |
Python3
# Python program to return first node of loopclass Node: def __init__(self,d): self.data = d self.next = Nonehead = Nonedef push(new_data): global head new_node = Node(new_data) new_node.next = head head=new_nodedef detectLoop(h): global head if (head == None): return False else: while (head != None): if (head.data == -1): return True else: head.data = -1 head = head.next return Falsepush(1);push(2);push(3);push(4);push(5);head.next.next.next.next.next = head.next.nextif (detectLoop(head)): print("1")else: print("0") # This code is contributed by patel2127. |
C#
// C# program to return first node of loopusing System;public class Node{ public int data; public Node next; public Node(int d) { data = d; next = null; }} public class GFG{ // Head of liststatic Node head; /* 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;} // Function to detect first node of loop// in a linked list that may contain loopstatic bool detectLoop(Node h){ // If the head is null we will return false if (head == null) return false; else { // Traversing the linked list // for detecting loop while (head != null) { // If loop found if (head.data == -1) { return true; } // Changing the data of visited node to any // value which is outside th given range // here it is supposed the given range is (1 // <= Data on Node <= 10^3) else { head.data = -1; head = head.next; } } // If loop not found return false return false; }} // Driver Code static public void Main (){ push(1); push(2); push(3); push(4); push(5); /* Create a loop for testing */ head.next.next.next.next.next = head.next.next; if (detectLoop(head)) Console.WriteLine("1"); else Console.WriteLine("0"); }} |
Javascript
<script>// Javascript program to return first node of loop// Linked list Nodeclass Node{ constructor(d) { this.data = d; this.next = null; }}// Head of listlet head;/* Inserts a new Node at front of the list. */function push(new_data){ /* 1 & 2: Allocate the Node & Put in the data*/ let 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;}// Function to detect first node of loop// in a linked list that may contain loopfunction detectLoop(h){ // If the head is null we will return false if (head == null) return false; else { // Traversing the linked list // for detecting loop while (head != null) { // If loop found if (head.data == -1) { return true; } // Changing the data of visited node to any // value which is outside th given range // here it is supposed the given range is (1 // <= Data on Node <= 10^3) else { head.data = -1; head = head.next; } } // If loop not found return false return false; }}// Driver Codepush(1);push(2);push(3);push(4);push(5);/* Create a loop for testing */head.next.next.next.next.next = head.next.next;if (detectLoop(head)) document.write("1");else document.write("0");// This code is contributed by unknown2108</script> |
Output
1
Time Complexity: O(N)
Auxiliary Space: O(1)
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