C++ program to find the intersection point of two linked lists

C++ program to find the intersection point of two linked lists


The idea is to get a pointer to the loop node using Floyd’s cycle detection algorithm  and count the total number of nodes in the loop using that loop node, say k. Then take two pointers – one pointing to the head node and another pointing to the kth node from the head. If we simultaneously move these pointers at the same speed, they will eventually meet at the loop’s starting node.


#include <iostream>

#include <unordered_set>

using namespace std;


// A Linked List Node

struct Node


    int data;

    Node* next;



// Utility function to create a new node with the given data and

// pushes it onto the list's front

void push(Node*& headRef, int data)


    // create a new linked list node from the heap

    Node* newNode = new Node;


    newNode->data = data;

    newNode->next = headRef;

    headRef = newNode;



// Find the starting node of the loop in a linked list pointed by `head`.

// The `loopNode` points to one of the nodes involved in the cycle

Node* removeCycle(Node* loopNode, Node* head)


    // find the count of nodes involved in the loop and store the count in `k`

    int k = 1;

    for (Node* ptr = loopNode; ptr->next != loopNode; ptr = ptr->next) {




    // get pointer to k'th node from the head

    Node* curr = head;

    for (int i = 0; i < k; i++) {

        curr = curr->next;



    // simultaneously move the `head` and `curr` pointers

    // at the same speed until they meet

    while (curr != head)


        curr = curr->next;

        head = head->next;



    // `curr` now points to the starting node of the loop

    return curr;



// Function to identify a cycle in a linked list using

// Floyd’s cycle detection algorithm

Node* identifyCycle(Node* head)


    // take two pointers – `slow` and `fast`

    Node *slow = head, *fast = head;


    while (fast && fast->next)


        // move slow by one pointer

        slow = slow->next;


        // move fast by two pointers

        fast = fast->next->next;


        // if they meet any node, the linked list contains a cycle

        if (slow == fast) {

            return slow;




    // return null if the linked list does not contain a cycle

    return nullptr;



// Function to find the intersection point of two linked lists

Node* findIntersection(Node* first, Node* second)


    Node* prev = nullptr;       // previous pointer

    Node* curr = first;         // main pointer


    // traverse the first list

    while (curr)


        // update the previous pointer to the current node and

        // move the main pointer to the next node

        prev = curr;

        curr = curr->next;



    // create a cycle in the first list

    if (prev) {

        prev->next = first;



    // now get a pointer to the loop node using the second list

    Node* slow = identifyCycle(second);


    // find the intersection node

    Node* addr = nullptr;

    if (slow) {

        addr = removeCycle(slow, second);



    // remove cycle in the first list before exiting

    if (prev) {

        prev->next = nullptr;



    // return the intersection node

    return addr;



int main()


    // construct the first linked list (1 —> 2 —> 3 —> 4 —> 5 —> null)

    Node* first = nullptr;

    for (int i = 7; i > 0; i--) {

        push(first, i);



    // construct the second linked list (1 —> 2 —> 3 —> null)

    Node* second = nullptr;

    for (int i = 3; i > 0; i--) {

        push(second, i);



    // link tail of the second list to the fourth node of the first list

    second->next->next->next = first->next->next->next;


    Node* addr = findIntersection(first, second);

    if (addr) {

        cout << "The intersection point is " << addr->data << endl;


    else {

        cout << "The lists do not intersect." << endl;



    return 0;



The intersection point is 4

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