Find length of loop in linked list
Write a function detectAndCountLoop() that checks whether a given Linked List contains loop and if loop is present then returns count of nodes in loop. For example, the loop is present in below-linked list and length of the loop is 4. If the loop is not present, then the function should return 0.
Approach: It is known that Floyd’s Cycle detection algorithm terminates when fast and slow pointers meet at a common point. It is also known that this common point is one of the loop nodes. Store the address of this common point in a pointer variable say (ptr). Then initialize a counter with 1 and start from the common point and keeps on visiting the next node and increasing the counter till the common pointer is reached again.
At that point, the value of the counter will be equal to the length of the loop.
Algorithm:
- Find the common point in the loop by using the Floyd’s Cycle detection algorithm
- Store the pointer in a temporary variable and keep a count = 0
- Traverse the linked list until the same node is reached again and increase the count while moving to next node.
- Print the count as length of loop
C++
// C++ program to count number of nodes// in loop in a linked list if loop is// present#include<bits/stdc++.h>using namespace std;/* Link list node */struct Node{ int data; struct Node* next;};// Returns count of nodes present in loop.int countNodes(struct Node *n){ int res = 1; struct Node *temp = n; while (temp->next != n) { res++; temp = temp->next; } return res;}/* This function detects and counts loopnodes in the list. If loop is not therein then returns 0 */int countNodesinLoop(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 and fast_p meet at some point then there is a loop */ if (slow_p == fast_p) return countNodes(slow_p); } /* Return 0 to indeciate that their is no loop*/ return 0;}struct Node *newNode(int key){ struct Node *temp = (struct Node*)malloc(sizeof(struct Node)); temp->data = key; temp->next = NULL; return temp;}// Driver Codeint main(){ struct 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 */ head->next->next->next->next->next = head->next; cout << countNodesinLoop(head) << endl; return 0;}// This code is contributed by SHUBHAMSINGH10 |
C
// C program to count number of nodes// in loop in a linked list if loop is// present#include<stdio.h>#include<stdlib.h>/* Link list node */struct Node{ int data; struct Node* next;};// Returns count of nodes present in loop.int countNodes(struct Node *n){ int res = 1; struct Node *temp = n; while (temp->next != n) { res++; temp = temp->next; } return res;}/* This function detects and counts loop nodes in the list. If loop is not there in then returns 0 */int countNodesinLoop(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 and fast_p meet at some point then there is a loop */ if (slow_p == fast_p) return countNodes(slow_p); } /* Return 0 to indeciate that ther is no loop*/ return 0;}struct Node *newNode(int key){ struct Node *temp = (struct Node*)malloc(sizeof(struct Node)); temp->data = key; temp->next = NULL; return temp;}/* Driver program to test above function*/int main(){ struct 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 */ head->next->next->next->next->next = head->next; printf("%d \n", countNodesinLoop(head)); return 0;} |
Java
// Java program to count number of nodes// in loop in a linked list if loop is// presentimport java.io.*;class GFG { /* Link list node */static class Node{ int data; Node next; Node(int data) { this.data =data; next =null; }}// Returns count of nodes present in loop.static int countNodes( Node n){int res = 1;Node temp = n;while (temp.next != n){ res++; temp = temp.next;}return res;}/* This function detects and counts loopnodes in the list. If loop is not therein then returns 0 */static int countNodesinLoop( Node list){ Node slow_p = list, fast_p = list; 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 and fast_p meet at some point then there is a loop */ if (slow_p == fast_p) return countNodes(slow_p); } /* Return 0 to indeciate that ther is no loop*/ return 0;}static Node newNode(int key){ Node temp = new Node(key); return temp;}/* 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 */ head.next.next.next.next.next = head.next; System.out.println( countNodesinLoop(head)); }}// This code is contributed by inder_verma. |
Python3
# Python 3 program to find the number# of nodes in looop in a linked list# if loop is present# Python Code to detect a loop and# find the length of the loop# Node defining classclass Node: # Function to make a node def __init__(self, val): self.val = val self.next = None # Linked List defining and loop# length finding classclass LinkedList: # Function to initialize the # head of the linked list def __init__(self): self.head = None # Function to insert a new # node at the end def AddNode(self, val): if self.head is None: self.head = Node(val) else: curr = self.head while(curr.next): curr = curr.next curr.next = Node(val) # Function to create a loop in the # Linked List. This function creates # a loop by connecting the last node # to n^th node of the linked list, # (counting first node as 1) def CreateLoop(self, n): # LoopNode is the connecting node to # the last node of linked list LoopNode = self.head for _ in range(1, n): LoopNode = LoopNode.next # end is the last node of the Linked List end = self.head while(end.next): end = end.next # Creating the loop end.next = LoopNode # Function to detect the loop and return # the length of the loop if the returned # value is zero, that means that either # the linked list is empty or the linked # list doesn't have any loop def detectLoop(self): # if linked list is empty then there # is no loop, so return 0 if self.head is None: return 0 # Using Floyd’s Cycle-Finding # Algorithm/ Slow-Fast Pointer Method slow = self.head fast = self.head flag = 0 # to show that both slow and fast # are at start of the Linked List while(slow and slow.next and fast and fast.next and fast.next.next): if slow == fast and flag != 0: # Means loop is confirmed in the # Linked List. Now slow and fast # are both at the same node which # is part of the loop count = 1 slow = slow.next while(slow != fast): slow = slow.next count += 1 return count slow = slow.next fast = fast.next.next flag = 1 return 0 # No loop # Setting up the code# Making a Linked List and adding the nodesmyLL = LinkedList()myLL.AddNode(1)myLL.AddNode(2)myLL.AddNode(3)myLL.AddNode(4)myLL.AddNode(5)# Creating a loop in the linked List# Loop is created by connecting the# last node of linked list to n^th node# 1<= n <= len(LinkedList)myLL.CreateLoop(2)# Checking for Loop in the Linked List# and printing the length of the looploopLength = myLL.detectLoop()if myLL.head is None: print("Linked list is empty")else: print(str(loopLength))# This code is contributed by _Ashutosh |
C#
// C# program to count number of nodes// in loop in a linked list if loop is// presentusing System;class GFG{ /* Link list node */ class Node { public int data; public Node next; public Node(int data) { this.data = data; next = null; } } // Returns count of nodes present in loop. static int countNodes( Node n) { int res = 1; Node temp = n; while (temp.next != n) { res++; temp = temp.next; } return res; } /* This function detects and counts loop nodes in the list. If loop is not there in then returns 0 */ static int countNodesinLoop( Node list) { Node slow_p = list, fast_p = list; 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 and fast_p meet at some point then there is a loop */ if (slow_p == fast_p) return countNodes(slow_p); } /* Return 0 to indeciate that ther is no loop*/ return 0; } static Node newNode(int key) { Node temp = new Node(key); return temp; } /* 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 */ head.next.next.next.next.next = head.next; Console.WriteLine( countNodesinLoop(head)); }}// This code is contributed by Rajput-Ji |
Javascript
<script>// javascript program to count number of nodes// in loop in a linked list if loop is// present /* Link list node */class Node { constructor(data) { this.data = data; this.next = null; }} // Returns count of nodes present in loop. function countNodes( n) { var res = 1; temp = n; while (temp.next != n) { res++; temp = temp.next; } return res; } /* * This function detects and counts loop nodes in the list. If loop is not there * in then returns 0 */ function countNodesinLoop( list) { var slow_p = list, fast_p = list; 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 and fast_p meet at some povar then there is a loop */ if (slow_p == fast_p) return countNodes(slow_p); } /* Return 0 to indeciate that ther is no loop */ return 0; } function newNode(key) { temp = new Node(key); return temp; } /* Driver program to test above function */ head = newNode(1); head.next = newNode(2); head.next.next = newNode(3); head.next.next.next = newNode(4); head.next.next.next.next = newNode(5); /* Create a loop for testing */ head.next.next.next.next.next = head.next; document.write(countNodesinLoop(head));// This code contributed by gauravrajput1</script> |
Output :
4
Complexity Analysis:
- Time complexity:O(n).
Only one traversal of the linked list is needed. - Auxiliary Space:O(1).
As no extra space is required.
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