Find length of loop/cycle in given Linked List

Given the head of a linked list. The task is to find if a loop exists in the linked list if yes then return the length of the loop in the linked list else return 0.

Examples:

Input: linked list =
Output: 4
Explanation: The loop is present in the below-linked list and the length of the loop is 4. 

Input: linked list = 4 -> 3 -> 7 -> 9 -> 2
Output: 0

Approach: Below is the idea to solve the problem:

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 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.

Follow the below steps to implement the idea:

• 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

Below is the implementation of the above approach:

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 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 indicate that
       there 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 Code
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;

    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 indicate that there 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
// present
import 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 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 indicate that there 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 loop in a linked list
# if loop is present

# Python Code to detect a loop and
# find the length of the loop
# Node defining class


class Node:

    # Function to make a node
    def __init__(self, val):
        self.val = val
        self.next = None

# Linked List defining and loop
# length finding class


class 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 nodes
myLL = 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 loop
loopLength = 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
// present
using 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 indicate that there 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 point then there is a loop
             */
            if (slow_p == fast_p)
                return countNodes(slow_p);
        }

        /* Return 0 to indicate that there 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

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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