Write a function findFirstLoopNode() that checks whether a given Linked List contains a loop. If the loop is present then it returns point to the first node of the loop. Else it returns NULL.
Example :
Input : Head of below linked list
Output : Pointer to node 2
We have discussed Floyd’s loop detection algorithm. Below are steps to find the first node of the loop.
1. If a loop is found, initialize a slow pointer to head, let fast pointer be at its position.
2. Move both slow and fast pointers one node at a time.
3. The point at which they meet is the start of the loop.
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 and remove loop// in a linked list that may contain loopNode* detectAndRemoveLoop(Node* head){ // If list is empty or has only one node // without loop if (head == NULL || head->next == NULL) return NULL; Node *slow = head, *fast = head; // Move slow and fast 1 and 2 steps // ahead respectively. slow = slow->next; fast = fast->next->next; // Search for loop using slow and // fast pointers while (fast && fast->next) { if (slow == fast) break; slow = slow->next; fast = fast->next->next; } // If loop does not exist if (slow != fast) return NULL; // If loop exists. Start slow from // head and fast from meeting point. slow = head; while (slow != fast) { slow = slow->next; fast = fast->next; } return slow;}/* Driver program to test above function*/int main(){ Node* head = newNode(50); head->next = newNode(20); head->next->next = newNode(15); head->next->next->next = newNode(4); head->next->next->next->next = newNode(10); /* Create a loop for testing */ head->next->next->next->next->next = head->next->next; Node* res = detectAndRemoveLoop(head); if (res == NULL) cout << "Loop does not exist"; else cout << "Loop starting node is " << res->key; return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
C
// C++ program to return first node of loop.#include <stdio.h>#include <stdlib.h>typedef struct Node { int key; struct Node* next;} Node;Node* newNode(int key){ Node* temp = (Node*)malloc(sizeof(Node)); temp->key = key; temp->next = NULL; return temp;}// A utility function to print a linked listvoid printList(Node* head){ while (head != NULL) { printf("%d ",head->key); head = head->next; } printf("/n");}// Function to detect and remove loop// in a linked list that may contain loopNode* detectAndRemoveLoop(Node* head){ // If list is empty or has only one node // without loop if (head == NULL || head->next == NULL) return NULL; Node *slow = head, *fast = head; // Move slow and fast 1 and 2 steps // ahead respectively. slow = slow->next; fast = fast->next->next; // Search for loop using slow and // fast pointers while (fast && fast->next) { if (slow == fast) break; slow = slow->next; fast = fast->next->next; } // If loop does not exist if (slow != fast) return NULL; // If loop exists. Start slow from // head and fast from meeting point. slow = head; while (slow != fast) { slow = slow->next; fast = fast->next; } return slow;}/* Driver program to test above function*/int main(){ Node* head = newNode(50); head->next = newNode(20); head->next->next = newNode(15); head->next->next->next = newNode(4); head->next->next->next->next = newNode(10); /* Create a loop for testing */ head->next->next->next->next->next = head->next->next; Node* res = detectAndRemoveLoop(head); if (res == NULL) printf("Loop does not exist"); else printf("Loop starting node is %d ",res->key); return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
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 liststatic void printList(Node head){ while (head != null) { System.out.print(head.key + " "); head = head.next; } System.out.println();}// Function to detect and remove loop// in a linked list that may contain loopstatic Node detectAndRemoveLoop(Node head){ // If list is empty or has // only one node without loop if (head == null || head.next == null) return null; Node slow = head, fast = head; // Move slow and fast 1 // and 2 steps ahead // respectively. slow = slow.next; fast = fast.next.next; // Search for loop using // slow and fast pointers while (fast != null && fast.next != null) { if (slow == fast) break; slow = slow.next; fast = fast.next.next; } // If loop does not exist if (slow != fast) return null; // If loop exists. Start slow from // head and fast from meeting point. slow = head; while (slow != fast) { slow = slow.next; fast = fast.next; } return slow;}// Driver codepublic static void main(String[] args){ Node head = newNode(50); head.next = newNode(20); head.next.next = newNode(15); head.next.next.next = newNode(4); head.next.next.next.next = newNode(10); // Create a loop for testing head.next.next.next.next.next = head.next.next; Node res = detectAndRemoveLoop(head); if (res == null) System.out.print("Loop does not exist"); else System.out.print("Loop starting node is " + res.key);}}// This code is contributed by shikhasingrajput |
Python3
# Python3 program to return first node of loop.class Node: def __init__(self, key): self.key = key self.next = None def newNode(key): temp = Node(key) return temp# A utility function to print a linked listdef printList(head): while (head != None): print(head.key, end = ' ') head = head.next print() # Function to detect and remove loop# in a linked list that may contain loopdef detectAndRemoveLoop(head): # If list is empty or has only one node # without loop if (head == None or head.next == None): return None slow = head fast = head # Move slow and fast 1 and 2 steps # ahead respectively. slow = slow.next fast = fast.next.next # Search for loop using slow and # fast pointers while (fast and fast.next): if (slow == fast): break slow = slow.next fast = fast.next.next # If loop does not exist if (slow != fast): return None # If loop exists. Start slow from # head and fast from meeting point. slow = head while (slow != fast): slow = slow.next fast = fast.next return slow# Driver codeif __name__=='__main__': head = newNode(50) head.next = newNode(20) head.next.next = newNode(15) head.next.next.next = newNode(4) head.next.next.next.next = newNode(10) # Create a loop for testing head.next.next.next.next.next = head.next.next res = detectAndRemoveLoop(head) if (res == None): print("Loop does not exist") else: print("Loop starting node is " + str(res.key)) # This code is contributed by rutvik_56 |
C#
// C# program to return // first node of loop.using System;class GFG{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 liststatic void printList(Node head){ while (head != null) { Console.Write(head.key + " "); head = head.next; } Console.WriteLine();}// Function to detect and remove loop// in a linked list that may contain loopstatic Node detectAndRemoveLoop(Node head){ // If list is empty or has // only one node without loop if (head == null || head.next == null) return null; Node slow = head, fast = head; // Move slow and fast 1 // and 2 steps ahead // respectively. slow = slow.next; fast = fast.next.next; // Search for loop using // slow and fast pointers while (fast != null && fast.next != null) { if (slow == fast) break; slow = slow.next; fast = fast.next.next; } // If loop does not exist if (slow != fast) return null; // If loop exists. Start slow from // head and fast from meeting point. slow = head; while (slow != fast) { slow = slow.next; fast = fast.next; } return slow;}// Driver codepublic static void Main(String[] args){ Node head = newNode(50); head.next = newNode(20); head.next.next = newNode(15); head.next.next.next = newNode(4); head.next.next.next.next = newNode(10); // Create a loop for testing head.next.next.next.next.next = head.next.next; Node res = detectAndRemoveLoop(head); if (res == null) Console.Write("Loop does not exist"); else Console.Write("Loop starting node is " + res.key);}}// This code is contributed by shikhasingrajput |
Javascript
<script>// Javascript program to return// first node of loop.class Node{ constructor(key) { this.key=key; this.next=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 and remove loop// in a linked list that may contain loopfunction detectAndRemoveLoop(head){ // If list is empty or has // only one node without loop if (head == null || head.next == null) return null; let slow = head, fast = head; // Move slow and fast 1 // and 2 steps ahead // respectively. slow = slow.next; fast = fast.next.next; // Search for loop using // slow and fast pointers while (fast != null && fast.next != null) { if (slow == fast) break; slow = slow.next; fast = fast.next.next; } // If loop does not exist if (slow != fast) return null; // If loop exists. Start slow from // head and fast from meeting point. slow = head; while (slow != fast) { slow = slow.next; fast = fast.next; } return slow;}// Driver codelet head = new Node(50); head.next = new Node(20); head.next.next = new Node(15); head.next.next.next = new Node(4); head.next.next.next.next = new Node(10); // Create a loop for testing head.next.next.next.next.next = head.next.next; let res = detectAndRemoveLoop(head); if (res == null) document.write("Loop does not exist"); else document.write("Loop starting node is " + res.key);// This code is contributed by unknown2108</script> |
Output:
Loop starting node is 15
Time Complexity: O(N), where N is length of array.
Auxiliary Space: O(1)
How does this approach work?
Let slow and fast meet at some point after Floyd’s Cycle finding algorithm. The below diagram shows the situation when the cycle is found.
We can conclude below from the above diagram
Distance traveled by fast pointer = 2 * (Distance traveled
by slow pointer)
(m + n*x + k) = 2*(m + n*y + k)
Note that before meeting the point shown above, fast
was moving at twice speed.
x --> Number of complete cyclic rounds made by
fast pointer before they meet first time
y --> Number of complete cyclic rounds made by
slow pointer before they meet first time
From the above equation, we can conclude below
m + k = (x-2y)*n Which means m+k is a multiple of n.
So if we start moving both pointers again at the same speed such that one pointer (say slow) begins from the head node of the linked list and other pointers (say fast) begins from the meeting point. When the slow pointer reaches the beginning of the loop (has made m steps), the fast pointer would have made also moved m steps as they are now moving the same pace. Since m+k is a multiple of n and fast starts from k, they would meet at the beginning. Can they meet before also? No, because the slow pointer enters the cycle first time after m steps.
Method 2:
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 return that node.
The code runs in O(n) time complexity and uses constant memory space.
Steps to solve this problem:
1. Create a temporary node and a pointer temp.
2. While head is not null:
*Check if head->next is null then return null.
*Check if head->next is temp than break.
*Create a nex pointer to a node and point it to head->next.
*Point head->next to temp.
*Update head to nex.
3. Return head.
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 loopNode* 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 NULL; } // Check if next is already // pointing to temp if (head->next == temp) { break; } // 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 head;}/* Driver program to test above function*/int main(){ Node* head = newNode(50); head->next = newNode(20); head->next->next = newNode(15); head->next->next->next = newNode(4); head->next->next->next->next = newNode(10); /* Create a loop for testing */ head->next->next->next->next->next = head->next->next; Node* res = detectLoop(head); if (res == NULL) cout << "Loop does not exist"; else cout << "Loop starting node is " << res->key; return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
C
// C program to return first node of loop#include <stdio.h>#include <stdlib.h>typedef struct Node { int key; struct Node* next;} Node;Node* newNode(int key){ Node* temp = (Node*)malloc(sizeof(Node)); temp->key = key; temp->next = NULL; return temp;}// A utility function to print a linked listvoid printList(Node* head){ while (head != NULL) { printf("%d ", head->key); head = head->next; } printf("\n");}// Function to detect first node of loop// in a linked list that may contain loopNode* detectLoop(Node* head){ // Create a temporary node Node* temp = (Node *)malloc(sizeof(Node)); while (head != NULL) { // This condition is for the case when there is no // loop if (head->next == NULL) return NULL; // Check if next is already pointing to temp if (head->next == temp) break; // 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 head;}/* Driver program to test above function*/int main(){ Node* head = newNode(50); head->next = newNode(20); head->next->next = newNode(15); head->next->next->next = newNode(4); head->next->next->next->next = newNode(10); /* Create a loop for testing */ head->next->next->next->next->next = head->next->next; Node* res = detectLoop(head); if (res == NULL) printf("Loop does not exist"); else printf("Loop starting node is %d ", res->key); return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
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();}// Function to detect first node of loop// in a linked list that may contain loopstatic Node 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 null; } // Check if next is already // pointing to temp if (head.next == temp) { break; } // 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 head;}/* Driver program to test above function*/public static void main(String[] args){ Node head = newNode(50); head.next = newNode(20); head.next.next = newNode(15); head.next.next.next = newNode(4); head.next.next.next.next = newNode(10); /* Create a loop for testing */ head.next.next.next.next.next = head.next.next; Node res = detectLoop(head); if (res == null) System.out.print("Loop does not exist"); else System.out.print("Loop starting node is " + res.key);}}// This code is contributed by gauravrajput1 |
Python3
# Python3 program to return first node of loopclass Node: def __init__(self, x): self.key = x self.next = None# A utility function to print a linked listdef printList(head): while (head != None): print(head.key, end = " ") head = head.next # Function to detect first node of loop# in a linked list that may contain loopdef detectLoop(head): # Create a temporary node temp = Node(-1) while (head != None): # This condition is for the case # when there is no loop if (head.next == None): return None # Check if next is already # pointing to temp if (head.next == temp): break # Store the pointer to the next node # in order to get to it in the next step nex = head.next # Make next point to temp head.next = temp # Get to the next node in the list head = nex return head# Driver codeif __name__ == '__main__': head = Node(50) head.next = Node(20) head.next.next = Node(15) head.next.next.next = Node(4) head.next.next.next.next = Node(10) # Create a loop for testing head.next.next.next.next.next = head.next.next res = detectLoop(head) if (res == None): print("Loop does not exist") else: print("Loop starting node is ", res.key)# This code is contributed by mohit kumar 29 |
C#
// C# program to return first node of loopusing System;class GFG{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 liststatic 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 loopstatic Node 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 null; } // Check if next is already // pointing to temp if (head.next == temp) { break; } // 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 head;}// Driver codepublic static void Main(String[] args){ Node head = newNode(50); head.next = newNode(20); head.next.next = newNode(15); head.next.next.next = newNode(4); head.next.next.next.next = newNode(10); // Create a loop for testing head.next.next.next.next.next = head.next.next; Node res = detectLoop(head); if (res == null) Console.Write("Loop does not exist"); else Console.Write("Loop starting node is " + res.key);}}// This code is contributed by Amit Katiyar |
Javascript
<script>// javascript program to return first node of loop class Node { constructor() { this.key = 0; this.next = null; } } function newNode(key) { temp = new Node(); temp.key = key; temp.next = null; return temp; } // A utility function to print a linked list function 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 loop function detectLoop( head) { // Create a temporary node temp = new Node(); while (head != null) { // This condition is for the case // when there is no loop if (head.next == null) { return null; } // Check if next is already // pointing to temp if (head.next == temp) { break; } // Store the pointer to the next node // in order to get to it in the next step nex = head.next; // Make next point to temp head.next = temp; // Get to the next node in the list head = nex; } return head; } /* Driver program to test above function */ head = newNode(50); head.next = newNode(20); head.next.next = newNode(15); head.next.next.next = newNode(4); head.next.next.next.next = newNode(10); /* Create a loop for testing */ head.next.next.next.next.next = head.next.next; res = detectLoop(head); if (res == null) document.write("Loop does not exist"); else document.write("Loop starting node is " + res.key);// This code contributed by gauravrajput1 </script> |
Output:
Loop starting node is 15
Time Complexity: O(N)
Auxiliary Space: O(1)
Method 3:
We can also use the concept of hashing in order to detect the first node of the loop. The idea is simple just iterate over the entire linked list and store node addresses in a set(C++ STL) one by one, while adding the node address into the set check if it already contains that particular node address if not then add node address to set if it is already present in the set then the current node is the first node of the loop.
C++14
// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002)// The below function take head of Linked List as// input and return Address of first node in// the loop if present else return NULL.#include<bits/stdc++.h>using namespace std;// Structure of linked list nodestruct ListNode{ int val; ListNode *next; ListNode(int x){ this->val = x; this->next = NULL; }};ListNode* detectCycle(ListNode* A){ // declaring map to store node address unordered_set<ListNode*> uset; ListNode *ptr = A; // Default consider that no cycle is present while (ptr != NULL) { // checking if address is already present in map if (uset.find(ptr) != uset.end()) return ptr; // if address not present then insert into the set else uset.insert(ptr); ptr = ptr->next; } return NULL;}/* Driver program to test above function*/int main(){ ListNode* head = new ListNode(50); head->next = new ListNode(20); head->next->next = new ListNode(15); head->next->next->next = new ListNode(4); head->next->next->next->next = new ListNode(10); // creating a loop for linked list head->next->next->next->next->next = head->next->next; ListNode* res = detectCycle(head); if (res == NULL) cout << "Loop does not exist"; else cout << "Loop starting node is " << res->val; return 0;} |
Java
// The below function take head of Linked List as// input and return Address of first node in// the loop if present else return NULL.import java.io.*;import java.util.*;class GFG{ static Node detectCycle(Node A) { // declaring map to store node address Set<Node> uset = new HashSet<Node>(); Node = A; // Default consider that no cycle is present while (ptr != NULL) { // checking if address is already present in map if(uset.contains(ptr)) { return ptr; } // if address not present then insert into the set else { uset.add(ptr); } ptr = ptr.next; } return null; }}// This code is contributed by avanitrachhadiya2155 |
Python3
# The below function take head of Linked List as# input and return Address of first node in# the loop if present else return NULL. ''' Definition for singly-linked list. * class ListNode: * def __init__(self, x): * self.val=x * self.next=None * '''def detectCycle(A): # Declaring map to store node # address uset = set() ptr = A # Default consider that no cycle # is present while (ptr != None): # Checking if address is already # present in map if (ptr in uset): return ptr # If address not present then # insert into the set else: uset.add(ptr) ptr = ptr.next return None# This code is contributed by pratham76 |
C#
// The below function take head of Linked List as// input and return Address of first node in// the loop if present else return NULL.using System.Collections.Generic;public class GFG{class Node { public int key; public Node next;}; static Node detectCycle(Node A) { // declaring map to store node address HashSet<Node> uset = new HashSet<Node>(); Node ptr = A; // Default consider that no cycle is present while (ptr != null) { // checking if address is already present in map if(uset.Contains(ptr)) { return ptr; } // if address not present then insert into the set else { uset.Add(ptr); } ptr = ptr.next; } return null; }}// This code is contributed by 29AjayKumar |
Javascript
<script>// The below function take head of Linked List as// input and return Address of first node in// the loop if present else return NULL. function detectCycle(A) { // declaring map to store node address let uset = new Set(); let ptr = A; // Default consider that no cycle is present while (ptr != NULL) { // checking if address is already present in map if(uset.has(ptr)) { return ptr; } // if address not present then insert into the set else { uset.add(ptr); } ptr = ptr.next; } return null; }// This code is contributed by patel2127</script> |
Output
Loop starting node is 15
Time Complexity: O(NlogN)
Auxiliary Space: O(N)
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