Find intersection of two linked lists using C++ program

In this article, we are going to see how to find intersection of two linked lists efficiently using merge sort?
By Radib Kar Last updated : August 01, 2023

Problem statement

Given two singly linked list, write a C++ program to find the intersection of two given singly linked list.

Consider the below example with sample input and output:

Let the first linked list be:
6 -> 5 -> 2 -> 9 -> NULL

Let the second linked list to be:
2 - 7 -> NULL

So there intersection is:
2 -> NULL

Brute force approach

One technique can be to traverse all the nodes of a linked list and check whether the traversed node is in the other linked list or not. Such an approach takes O(m*n) times complexity. m, n=length of linked lists.

Efficient approach

Efficient approach use to use merge sort.

Sort both the linked list using merge sort. (for detailed refer to: Merge sort for single linked lists)
Scan each linked list and build intersection according to following:

IF (list1->data < list2->data)
	No intersection found
	Traverse list1 by one step( list1=list1->next)
ELSE IF(list1->data ==list2->data)
	Createnode(list1->data) && append node to the intersection list
	Traverse list1 by one step( list1=list1->next)
	Traverse list2 by one step( list2=list2->next)
ELSE//list1->data >list2->data
	No intersection found
	Traverse list2 by one step( list2=list2->next)
END IF-ELSE

Display the intersection list

Find intersection of two linked lists using C++ program

#include <bits/stdc++.h>
using namespace std;

class node {
public:
    int data; // data field
    struct node* next;
};

void display(class node* head)
{
    node* current = head; // current node set to head
    //traverse until current node isn't NULL
    while (current != NULL) {
        printf("%d ", current->data);
        current = current->next; // go to next node
    }
}

node* creatnode(int d)
{
    node* temp = (node*)malloc(sizeof(node));
    temp->data = d;
    temp->next = NULL;
    return temp;
}

//merging two sorted list
node* mergeList(node* split1, node* split2)
{
    node* newhead = NULL;
    //base cases
    if (split1 == NULL)
        return split2;
    if (split2 == NULL)
        return split1;

    if (split1->data <= split2->data) {
        newhead = split1;
        newhead->next = mergeList(split1->next, split2);
    }
    else {
        newhead = split2;
        newhead->next = mergeList(split1, split2->next);
    }
    return newhead;
}

//split list for merge sort
void splitList(node* head, node** split1, node** split2)
{
    node* slow = head;
    node* fast = head->next;

    while (fast != NULL) {
        fast = fast->next;
        if (fast != NULL) {
            slow = slow->next;
            fast = fast->next;
        }
    }

    *split1 = head;
    *split2 = slow->next;
    //spliting
    slow->next = NULL;
}

//merge sort
void mergeSort(node** refToHead)
{
    node* head = *refToHead;
    node *split1, *split2;
    //base case
    if (head == NULL || head->next == NULL) {
        return;
    }
    //split the list in two halves
    splitList(head, &split1, &split2);
    //recursively sort the two halves
    mergeSort(&split1);
    mergeSort(&split2);

    *refToHead = mergeList(split1, split2);

    return;
}

node* findIntersection(node* head1, node* head2)
{
    //base case
    if (head1 == NULL && head2 == NULL)
        return NULL;

    node *head4 = NULL, *temp;
    //for inserting the first common node
    while (head1 != NULL && head2 != NULL && head4 == NULL) {
        if (head1->data < head2->data) {
            head1 = head1->next;
        }
        //intersection nodes(intersection)
        else if (head1->data == head2->data) {
            head4 = creatnode(head1->data);
            temp = head4;
            head1 = head1->next;
            head2 = head2->next;
        }
        else {
            head2 = head2->next;
        }
    }

    //for other common nodes(intersection)
    while (head1 != NULL && head2 != NULL) {
        if (head1->data < head2->data) {
            head1 = head1->next;
        }
        //intersection nodes
        else if (head1->data == head2->data) {
            temp->next = creatnode(head1->data);
            temp = temp->next;

            head1 = head1->next;
            head2 = head2->next;
        }
        else {
            head2 = head2->next;
        }
    }

    return head4;
}

// Main code
int main()
{
    printf("creating the linked list by inserting new nodes at the end\n");
    printf("enter 0 to stop building the list, else enter any integer\n");
    
    int k;
    node *curr, *temp;
    
    cout << "enter list1...\n";
    scanf("%d", &k);
    node* head1 = creatnode(k); //buliding list, first node
    scanf("%d", &k);
    temp = head1;

    ///////////////////inserting at the end//////////////////////
    while (k) {
        curr = creatnode(k);
        temp->next = curr; //appending each node
        temp = temp->next;
        scanf("%d", &k);
    }
    
    cout << "displaying list1...\n";
    display(head1); // displaying the list
    
    cout << "\nenter list2...\n";
    scanf("%d", &k);
    node* head2 = creatnode(k); //buliding list, first node
    scanf("%d", &k);
    temp = head2;

    ///////////////////inserting at the end//////////////////////
    while (k) {
        curr = creatnode(k);
        temp->next = curr; //appending each node
        temp = temp->next;
        scanf("%d", &k);
    }
    
    cout << "displaying list1...\n";
    display(head2);

    //sort both the lists
    mergeSort(&head1);
    mergeSort(&head2);

    cout << "\ndisplaying their intersection...\n";

    node* head4 = findIntersection(head1, head2);
    if (head4)
        display(head4);
    else
        cout << "No intersection found\n";

    return 0;
}

Output

First run:
creating the linked list by inserting new nodes at the end
enter 0 to stop building the list, else enter any integer
enter list1...
6 5 2 9 0
displaying list1...
6 5 2 9
enter list2...
2 7 0
displaying list1...
2 7
displaying their intersection...
2 

Second run:
creating the linked list by inserting new nodes at the end
enter 0 to stop building the list, else enter any integer
enter list1...
6 7 8 0
displaying list1...
6 7 8
enter list2...
1 5 2 0
displaying list1...
1 5 2
displaying their intersection...
No intersection found

Example with explanation

First linked list: 6->5->2->9->NULL
Second linked list: 2->7->NULL

After applying merge sort:
First linked list: 2->5->6->9->NULL
Second linked list: 2->7->NULL
------------------------------------------------------
//for better understanding nodes are represented 
//only by respective values
head1=2
head2=2
FindIntersection(2,2):

0^th iteration
head1!=NULL && head2!=NULL
head1->data==head2->data //=2
thus head4(head of intersection list) =2
temp=head4=2
intersection list up to now:
2->NULL
head1=2->next=5;
head2=2->next=7;

1^st iteration
head1!=NULL && head2!=NULL
head1->data<head2->data //5<7
thushead1=5->next=6;
temp=same as before
intersection list up to now:
2->NULL
head2=same as before;

2^nd iteration
head1!=NULL && head2!=NULL
thushead1=6->next=9;
temp=same as before
intersection list up to now:
2->NULL
head2=same as before;

3^rd iteration
head1!=NULL && head2!=NULL
head1->data>head2->data //9>7
thushead2=7->next=NULL;
temp=same as before
intersection list up to now:
2->NULL
Head1=same as before;

4^th iteration
Head2=NULL
So, iteration stops & intersection list is build:
2->NULL