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Write a function to get the intersection point of two Linked Lists

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  • Difficulty Level : Medium
  • Last Updated : 05 Jul, 2022

There are two singly linked lists in a system. By some programming error, the end node of one of the linked lists got linked to the second list, forming an inverted Y-shaped list. Write a program to get the point where two linked lists merge. 

Y ShapedLinked List

The above diagram shows an example with two linked lists having 15 as intersection points.

Method 1(Simply use two loops):
Use 2 nested for loops. The outer loop will be for each node of the 1st list and the inner loop will be for the 2nd list. In the inner loop, check if any of the nodes of the 2nd list is the same as the current node of the first linked list. The time complexity of this method will be O(M * N) where m and n are the numbers of nodes in two lists.

Below is the code for the above approach:

C++




// C++ program to get intersection point of two linked list
#include <bits/stdc++.h>
using namespace std;
 
/* Link list node */
class Node {
public:
    int data;
    Node* next;
};
 
/* function to get the intersection point of two linked
lists head1 and head2 */
Node* getIntesectionNode(Node* head1, Node* head2)
{
    while (head2) {
        Node* temp = head1;
        while (temp) {
            // if both Nodes are same
            if (temp == head2)
                return head2;
            temp = temp->next;
        }
        head2 = head2->next;
    }
    // intersection is not present between the lists
    return NULL;
}
 
// Driver Code
int main()
{
    /*
        Create two linked lists
 
        1st 3->6->9->15->30
        2nd 10->15->30
 
        15 is the intersection point
    */
 
    Node* newNode;
 
    // Addition of new nodes
    Node* head1 = new Node();
    head1->data = 10;
 
    Node* head2 = new Node();
    head2->data = 3;
 
    newNode = new Node();
    newNode->data = 6;
    head2->next = newNode;
 
    newNode = new Node();
    newNode->data = 9;
    head2->next->next = newNode;
 
    newNode = new Node();
    newNode->data = 15;
    head1->next = newNode;
    head2->next->next->next = newNode;
 
    newNode = new Node();
    newNode->data = 30;
    head1->next->next = newNode;
 
    head1->next->next->next = NULL;
 
    Node* intersectionPoint
        = getIntesectionNode(head1, head2);
 
    if (!intersectionPoint)
        cout << " No Intersection Point \n";
    else
        cout << "Intersection Point: "
             << intersectionPoint->data << endl;
}
 
// This code is contributed by Tapesh(tapeshdua420)


Java




// Java Program to get intersection point of two linked
// lists.
 
class GFG {
 
    static class Node {
        int data;
        Node next;
        Node(int d)
        {
            data = d;
            next = null;
        }
    }
 
    /* function to get the intersection point of two linked
    lists head1 and head2 */
    public Node getIntersectionNode(Node head1, Node head2)
    {
        while (head2 != null) {
            Node temp = head1;
            while (temp != null) {
                // if both Nodes are same
                if (temp == head2) {
                    return head2;
                }
                temp = temp.next;
            }
            head2 = head2.next;
        }
        // If intersection is not present between the lists,
        // return NULL.
        return null;
    }
 
    public static void main(String[] args)
    {
        GFG list = new GFG();
 
        Node head1, head2;
 
        /*
                Create two linked lists
 
                1st 3->6->9->15->30
                2nd 10->15->30
 
                15 is the intersection point
        */
 
        head1 = new Node(10);
        head2 = new Node(3);
 
        Node newNode = new Node(6);
        head2.next = newNode;
 
        newNode = new Node(9);
        head2.next.next = newNode;
 
        newNode = new Node(15);
        head1.next = newNode;
        head2.next.next.next = newNode;
 
        newNode = new Node(30);
        head1.next.next = newNode;
 
        head1.next.next.next = null;
 
        Node intersectionPoint
            = list.getIntersectionNode(head1, head2);
 
        if (intersectionPoint == null) {
            System.out.print(" No Intersection Point \n");
        }
        else {
            System.out.print("Intersection Point: "
                             + intersectionPoint.data);
        }
    }
}
 
// This code is contributed by lokesh (lokeshmvs21).


Output

Intersection Point: 15

Method 2 (Mark Visited Nodes):
This solution requires modifications to the basic linked list data structure. Have a visited flag with each node. Traverse the first linked list and keep marking visited nodes. Now traverse the second linked list, If you see a visited node again then there is an intersection point, return the intersecting node. This solution works in O(m+n) but requires additional information with each node. A variation of this solution that doesn’t require modification to the basic data structure can be implemented using a hash. Traverse the first linked list and store the addresses of visited nodes in a hash. Now traverse the second linked list and if you see an address that already exists in the hash then return the intersecting node.

Method 3(Using the difference in node counts) 

  • Get the count of the nodes in the first list, let the count be c1.
  • Get the count of the nodes in the second list, let the count be c2.
  • Get the difference of counts d = abs(c1 – c2)
  • Now traverse the bigger list from the first node to d nodes so that from here onwards both the lists have an equal no of nodes
  • Then we can traverse both lists in parallel till we come across a common node. (Note that getting a common node is done by comparing the address of the nodes)

Below image is a dry run of the above approach:

Below is the implementation of the above approach :

C++




// C++ program to get intersection point of two linked list
#include <bits/stdc++.h>
using namespace std;
 
/* Link list node */
class Node {
public:
    int data;
    Node* next;
};
 
/* Function to get the counts of node in a linked list */
int getCount(Node* head);
 
/* function to get the intersection point of two linked
lists head1 and head2 where head1 has d more nodes than
head2 */
int _getIntesectionNode(int d, Node* head1, Node* head2);
 
/* function to get the intersection point of two linked
lists head1 and head2 */
int getIntesectionNode(Node* head1, Node* head2)
{
 
    // Count the number of nodes in
    // both the linked list
    int c1 = getCount(head1);
    int c2 = getCount(head2);
    int d;
 
    // If first is greater
    if (c1 > c2) {
        d = c1 - c2;
        return _getIntesectionNode(d, head1, head2);
    }
    else {
        d = c2 - c1;
        return _getIntesectionNode(d, head2, head1);
    }
}
 
/* function to get the intersection point of two linked
lists head1 and head2 where head1 has d more nodes than
head2 */
int _getIntesectionNode(int d, Node* head1, Node* head2)
{
    // Stand at the starting of the bigger list
    Node* current1 = head1;
    Node* current2 = head2;
 
    // Move the pointer forward
    for (int i = 0; i < d; i++) {
        if (current1 == NULL) {
            return -1;
        }
        current1 = current1->next;
    }
 
    // Move both pointers of both list till they
    // intersect with each other
    while (current1 != NULL && current2 != NULL) {
        if (current1 == current2)
            return current1->data;
 
        // Move both the pointers forward
        current1 = current1->next;
        current2 = current2->next;
    }
 
    return -1;
}
 
/* Takes head pointer of the linked list and
returns the count of nodes in the list */
int getCount(Node* head)
{
    Node* current = head;
 
    // Counter to store count of nodes
    int count = 0;
 
    // Iterate till NULL
    while (current != NULL) {
 
        // Increase the counter
        count++;
 
        // Move the Node ahead
        current = current->next;
    }
 
    return count;
}
 
// Driver Code
int main()
{
    /*
        Create two linked lists
     
        1st 3->6->9->15->30
        2nd 10->15->30
     
        15 is the intersection point
    */
 
    Node* newNode;
 
    // Addition of new nodes
    Node* head1 = new Node();
    head1->data = 10;
 
    Node* head2 = new Node();
    head2->data = 3;
 
    newNode = new Node();
    newNode->data = 6;
    head2->next = newNode;
 
    newNode = new Node();
    newNode->data = 9;
    head2->next->next = newNode;
 
    newNode = new Node();
    newNode->data = 15;
    head1->next = newNode;
    head2->next->next->next = newNode;
 
    newNode = new Node();
    newNode->data = 30;
    head1->next->next = newNode;
 
    head1->next->next->next = NULL;
 
    cout << "The node of intersection is " << getIntesectionNode(head1, head2);
}
 
// This code is contributed by rathbhupendra


C




// C program to get intersection point of two linked list
#include <stdio.h>
#include <stdlib.h>
 
/* Link list node */
struct Node {
    int data;
    struct Node* next;
};
 
/* Function to get the counts of node in a linked list */
int getCount(struct Node* head);
 
/* function to get the intersection point of two linked
   lists head1 and head2 where head1 has d more nodes than
   head2 */
int _getIntesectionNode(int d, struct Node* head1, struct Node* head2);
 
/* function to get the intersection point of two linked
   lists head1 and head2 */
int getIntesectionNode(struct Node* head1, struct Node* head2)
{
    int c1 = getCount(head1);
    int c2 = getCount(head2);
    int d;
 
    if (c1 > c2) {
        d = c1 - c2;
        return _getIntesectionNode(d, head1, head2);
    }
    else {
        d = c2 - c1;
        return _getIntesectionNode(d, head2, head1);
    }
}
 
/* function to get the intersection point of two linked
   lists head1 and head2 where head1 has d more nodes than
   head2 */
int _getIntesectionNode(int d, struct Node* head1, struct Node* head2)
{
    int i;
    struct Node* current1 = head1;
    struct Node* current2 = head2;
 
    for (i = 0; i < d; i++) {
        if (current1 == NULL) {
            return -1;
        }
        current1 = current1->next;
    }
 
    while (current1 != NULL && current2 != NULL) {
        if (current1 == current2)
            return current1->data;
        current1 = current1->next;
        current2 = current2->next;
    }
 
    return -1;
}
 
/* Takes head pointer of the linked list and
   returns the count of nodes in the list */
int getCount(struct Node* head)
{
    struct Node* current = head;
    int count = 0;
 
    while (current != NULL) {
        count++;
        current = current->next;
    }
 
    return count;
}
 
/* IGNORE THE BELOW LINES OF CODE. THESE LINES
   ARE JUST TO QUICKLY TEST THE ABOVE FUNCTION */
int main()
{
    /*
    Create two linked lists
 
    1st 3->6->9->15->30
    2nd 10->15->30
 
    15 is the intersection point
  */
 
    struct Node* newNode;
    struct Node* head1 = (struct Node*)malloc(sizeof(struct Node));
    head1->data = 10;
 
    struct Node* head2 = (struct Node*)malloc(sizeof(struct Node));
    head2->data = 3;
 
    newNode = (struct Node*)malloc(sizeof(struct Node));
    newNode->data = 6;
    head2->next = newNode;
 
    newNode = (struct Node*)malloc(sizeof(struct Node));
    newNode->data = 9;
    head2->next->next = newNode;
 
    newNode = (struct Node*)malloc(sizeof(struct Node));
    newNode->data = 15;
    head1->next = newNode;
    head2->next->next->next = newNode;
 
    newNode = (struct Node*)malloc(sizeof(struct Node));
    newNode->data = 30;
    head1->next->next = newNode;
 
    head1->next->next->next = NULL;
 
    printf("\n The node of intersection is %d \n",
           getIntesectionNode(head1, head2));
 
    getchar();
}


Java




// Java program to get intersection point of two linked list
 
class LinkedList {
 
    static Node head1, head2;
 
    static class Node {
 
        int data;
        Node next;
 
        Node(int d)
        {
            data = d;
            next = null;
        }
    }
 
    /*function to get the intersection point of two linked
    lists head1 and head2 */
    int getNode()
    {
        int c1 = getCount(head1);
        int c2 = getCount(head2);
        int d;
 
        if (c1 > c2) {
            d = c1 - c2;
            return _getIntesectionNode(d, head1, head2);
        }
        else {
            d = c2 - c1;
            return _getIntesectionNode(d, head2, head1);
        }
    }
 
    /* function to get the intersection point of two linked
     lists head1 and head2 where head1 has d more nodes than
     head2 */
    int _getIntesectionNode(int d, Node node1, Node node2)
    {
        int i;
        Node current1 = node1;
        Node current2 = node2;
        for (i = 0; i < d; i++) {
            if (current1 == null) {
                return -1;
            }
            current1 = current1.next;
        }
        while (current1 != null && current2 != null) {
            if (current1.data == current2.data) {
                return current1.data;
            }
            current1 = current1.next;
            current2 = current2.next;
        }
 
        return -1;
    }
 
    /*Takes head pointer of the linked list and
    returns the count of nodes in the list */
    int getCount(Node node)
    {
        Node current = node;
        int count = 0;
 
        while (current != null) {
            count++;
            current = current.next;
        }
 
        return count;
    }
 
    public static void main(String[] args)
    {
        LinkedList list = new LinkedList();
 
        // creating first linked list
        list.head1 = new Node(3);
        list.head1.next = new Node(6);
        list.head1.next.next = new Node(9);
        list.head1.next.next.next = new Node(15);
        list.head1.next.next.next.next = new Node(30);
 
        // creating second linked list
        list.head2 = new Node(10);
        list.head2.next = new Node(15);
        list.head2.next.next = new Node(30);
 
        System.out.println("The node of intersection is " + list.getNode());
    }
}
 
// This code has been contributed by Mayank Jaiswal


Python3




# defining a node for LinkedList
class Node:
  def __init__(self,data):
    self.data=data
    self.next=None
     
 
 
def getIntersectionNode(head1,head2):
   
  #finding the total number of elements in head1 LinkedList
    c1=getCount(head1)
   
  #finding the total number of elements in head2 LinkedList
    c2=getCount(head2)
   
  #Traverse the bigger node by 'd' so that from that node onwards, both LinkedList
  #would be having same number of nodes and we can traverse them together.
    if c1 > c2:
        d=c1-c2
        return _getIntersectionNode(d,head1,head2)
    else:
        d=c2-c1
        return _getIntersectionNode(d,head2,head1)
   
   
def _getIntersectionNode(d,head1,head2):
     
     
    current1=head1
    current2=head2
     
     
    for i in range(d):
        if current1 is None:
            return -1
        current1=current1.next
     
    while current1 is not None and current2 is not None:
     
    # Instead of values, we need to check if there addresses are same
    # because there can be a case where value is same but that value is
    #not an intersecting point.
        if current1 is current2:
            return current1.data # or current2.data ( the value would be same)
     
        current1=current1.next
        current2=current2.next
   
  # Incase, we are not able to find our intersecting point.
    return -1
   
#Function to get the count of a LinkedList
def getCount(node):
    cur=node
    count=0
    while cur is not None:
        count+=1
        cur=cur.next
    return count
     
 
if __name__ == '__main__':
  # Creating two LinkedList
  # 1st one: 3->6->9->15->30
  # 2nd one: 10->15->30
  # We can see that 15 would be our intersection point
   
  # Defining the common node
   
  common=Node(15)
   
  #Defining first LinkedList
   
  head1=Node(3)
  head1.next=Node(6)
  head1.next.next=Node(9)
  head1.next.next.next=common
  head1.next.next.next.next=Node(30)
   
  # Defining second LinkedList
   
  head2=Node(10)
  head2.next=common
  head2.next.next=Node(30)
   
  print("The node of intersection is ",getIntersectionNode(head1,head2))
   
  # The code is contributed by Ansh Gupta.


C#




// C# program to get intersection point of two linked list
using System;
class LinkedList {
 
    Node head1, head2;
 
    public class Node {
 
        public int data;
        public Node next;
 
        public Node(int d)
        {
            data = d;
            next = null;
        }
    }
 
    /*function to get the intersection point of two linked
    lists head1 and head2 */
    int getNode()
    {
        int c1 = getCount(head1);
        int c2 = getCount(head2);
        int d;
 
        if (c1 > c2) {
            d = c1 - c2;
            return _getIntesectionNode(d, head1, head2);
        }
        else {
            d = c2 - c1;
            return _getIntesectionNode(d, head2, head1);
        }
    }
 
    /* function to get the intersection point of two linked
    lists head1 and head2 where head1 has d more nodes than
    head2 */
    int _getIntesectionNode(int d, Node node1, Node node2)
    {
        int i;
        Node current1 = node1;
        Node current2 = node2;
        for (i = 0; i < d; i++) {
            if (current1 == null) {
                return -1;
            }
            current1 = current1.next;
        }
        while (current1 != null && current2 != null) {
            if (current1.data == current2.data) {
                return current1.data;
            }
            current1 = current1.next;
            current2 = current2.next;
        }
 
        return -1;
    }
 
    /*Takes head pointer of the linked list and
    returns the count of nodes in the list */
    int getCount(Node node)
    {
        Node current = node;
        int count = 0;
 
        while (current != null) {
            count++;
            current = current.next;
        }
 
        return count;
    }
 
    public static void Main(String[] args)
    {
        LinkedList list = new LinkedList();
 
        // creating first linked list
        list.head1 = new Node(3);
        list.head1.next = new Node(6);
        list.head1.next.next = new Node(9);
        list.head1.next.next.next = new Node(15);
        list.head1.next.next.next.next = new Node(30);
 
        // creating second linked list
        list.head2 = new Node(10);
        list.head2.next = new Node(15);
        list.head2.next.next = new Node(30);
 
        Console.WriteLine("The node of intersection is " + list.getNode());
    }
}
 
// This code is contributed by Arnab Kundu


Javascript




<script>
 
 
class Node
{
    constructor(item)
    {
        this.data=item;
        this.next=null;
    }
}
 
let head1,head2;
function getNode()
{
    let c1 = getCount(head1);
        let c2 = getCount(head2);
         
        let d;
   
        if (c1 > c2) {
            d = c1 - c2;
            return _getIntesectionNode(d, head1, head2);
        }
        else {
            d = c2 - c1;
            return _getIntesectionNode(d, head2, head1);
        }
}
 
function _getIntesectionNode(d,node1,node2)
{
    let i;
        let current1 = node1;
        let current2 = node2;
        for (i = 0; i < d; i++) {
            if (current1 == null) {
                return -1;
            }
            current1 = current1.next;
        }
        while (current1 != null && current2 != null) {
            if (current1.data == current2.data) {
                return current1.data;
            }
            current1 = current1.next;
            current2 = current2.next;
        }
   
        return -1;
}
 
function getCount(node)
{
    let current = node;
        let count = 0;
   
        while (current != null) {
            count++;
            current = current.next;
        }
   
        return count;
}
 
head1 = new Node(3);
head1.next = new Node(6);
head1.next.next = new Node(9);
head1.next.next.next = new Node(15);
head1.next.next.next.next = new Node(30);
 
// creating second linked list
head2 = new Node(10);
head2.next = new Node(15);
head2.next.next = new Node(30);
 
document.write("The node of intersection is " + getNode());
 
 
 
// This code is contributed by avanitrachhadiya2155
 
</script>


Output

The node of intersection is 15

Time Complexity: O(m+n) 
Auxiliary Space: O(1)

Method 4(Make a circle in the first list) 
Thanks to Saravanan Man for providing the below solution. 
1. Traverse the first linked list(count the elements) and make a circular linked list. (Remember the last node so that we can break the circle later on). 
2. Now view the problem as finding the loop in the second linked list. So the problem is solved. 
3. Since we already know the length of the loop(size of the first linked list) we can traverse those many numbers of nodes in the second list, and then start another pointer from the beginning of the second list. we have to traverse until they are equal, and that is the required intersection point. 
4. remove the circle from the linked list. 

Time Complexity: O(m+n) 
Auxiliary Space: O(1)

Method 5 (Reverse the first list and make equations) 
Thanks to Saravanan Mani for providing this method.  

1) Let X be the length of the first linked list until intersection point.
   Let Y be the length of the second linked list until the intersection point.
   Let Z be the length of the linked list from the intersection point to End of
   the linked list including the intersection node.
   We Have
           X + Z = C1;
           Y + Z = C2;
2) Reverse first linked list.
3) Traverse Second linked list. Let C3 be the length of second list - 1. 
     Now we have
        X + Y = C3
     We have 3 linear equations. By solving them, we get
       X = (C1 + C3 – C2)/2;
       Y = (C2 + C3 – C1)/2;
       Z = (C1 + C2 – C3)/2;
      WE GOT THE INTERSECTION POINT.
4)  Reverse first linked list.

Advantage: No Comparison of pointers. 
Disadvantage: Modifying linked list(Reversing list). 
Time complexity: O(m+n) 
Auxiliary Space: O(1)

Method 6 (Traverse both lists and compare addresses of last nodes) This method is only to detect if there is an intersection point or not. (Thanks to NeoTheSaviour for suggesting this)  

1) Traverse the list 1, store the last node address
2) Traverse the list 2, store the last node address.
3) If nodes stored in 1 and 2 are same then they are intersecting.

The time complexity of this method is O(m+n) and the used Auxiliary space is O(1)

Method 7 (Use Hashing) 
Basically, we need to find a common node of two linked lists. So we hash all nodes of the first list and then check the second list. 
1) Create an empty hash set. 
2) Traverse the first linked list and insert all nodes’ addresses in the hash set. 
3) Traverse the second list. For every node check if it is present in the hash set. If we find a node in the hash set, return the node.

C++




// C++ program to get intersection point of two linked list
#include <iostream>
#include <unordered_set>
using namespace std;
class Node
{
public:
    int data;
    Node* next;
      Node(int d)
    {
      data = d;
      next = NULL;
    }
};
 
// function to find the intersection point of two lists
void MegeNode(Node* n1, Node* n2)
{
    unordered_set<Node*> hs;
    while (n1 != NULL) {
        hs.insert(n1);
        n1 = n1->next;
    }
    while (n2) {
        if (hs.find(n2) != hs.end()) {
            cout << n2->data << endl;
            break;
        }
        n2 = n2->next;
    }
}
 
// function to print the list
void Print(Node* n)
{
      Node* curr = n;
      while(curr != NULL){
      cout << curr->data << " ";
      curr = curr->next;
    }
      cout << endl;
}
   
int main()
{
     // list 1
     Node* n1 = new Node(1);
     n1->next = new Node(2);
     n1->next->next = new Node(3);
     n1->next->next->next = new Node(4);
     n1->next->next->next->next = new Node(5);
     n1->next->next->next->next->next = new Node(6);
     n1->next->next->next->next->next->next = new Node(7);
     // list 2
     Node* n2 = new Node(10);
     n2->next = new Node(9);
     n2->next->next = new Node(8);
     n2->next->next->next = n1->next->next->next;
     Print(n1);
     Print(n2);
           
       MegeNode(n1,n2);
       
       return 0;
}
 
 // This code is contributed by Upendra


Java




// Java program to get intersection point of two linked list
import java.util.*;
class Node {
    int data;
    Node next;
    Node(int d)
    {
        data = d;
        next = null;
    }
}
class LinkedListIntersect {
    public static void main(String[] args)
    {
        // list 1
        Node n1 = new Node(1);
        n1.next = new Node(2);
        n1.next.next = new Node(3);
        n1.next.next.next = new Node(4);
        n1.next.next.next.next = new Node(5);
        n1.next.next.next.next.next = new Node(6);
        n1.next.next.next.next.next.next = new Node(7);
        // list 2
        Node n2 = new Node(10);
        n2.next = new Node(9);
        n2.next.next = new Node(8);
        n2.next.next.next = n1.next.next.next;
        Print(n1);
        Print(n2);
        System.out.println(MegeNode(n1, n2).data);
    }
 
    // function to print the list
    public static void Print(Node n)
    {
        Node cur = n;
        while (cur != null) {
            System.out.print(cur.data + "  ");
            cur = cur.next;
        }
        System.out.println();
    }
 
    // function to find the intersection of two node
    public static Node MegeNode(Node n1, Node n2)
    {
        // define hashset
        HashSet<Node> hs = new HashSet<Node>();
        while (n1 != null) {
            hs.add(n1);
            n1 = n1.next;
        }
        while (n2 != null) {
            if (hs.contains(n2)) {
                return n2;
            }
            n2 = n2.next;
        }
        return null;
    }
}


Python3




# Python program to get intersection
# point of two linked list
class Node :
    def __init__(self, d):
        self.data = d;
        self.next = None;
 
# Function to print the list
def Print(n):
    cur = n;
    while (cur != None) :
        print(cur.data, end=" ");
        cur = cur.next;
    print("");
 
# Function to find the intersection of two node
def MegeNode(n1, n2):
     
    # Define hashset
    hs = set();
 
    while (n1 != None):
        hs.add(n1);
        n1 = n1.next;
    while (n2 != None):
        if (n2 in hs):
            return n2;
        n2 = n2.next;
     
    return None;
 
 
# Driver code
 
# list 1
n1 = Node(1);
n1.next = Node(2);
n1.next.next = Node(3);
n1.next.next.next = Node(4);
n1.next.next.next.next = Node(5);
n1.next.next.next.next.next = Node(6);
n1.next.next.next.next.next.next = Node(7);
 
# list 2
n2 = Node(10);
n2.next = Node(9);
n2.next.next = Node(8);
n2.next.next.next = n1.next.next.next;
 
Print(n1);
Print(n2);
 
print(MegeNode(n1, n2).data);
 
# This code is contributed by _saurabh_jaiswal


C#




// C# program to get intersection point of two linked list
using System;
using System.Collections.Generic;
 
public class Node
{
    public int data;
    public Node next;
    public Node(int d)
    {
        data = d;
        next = null;
    }
}
public class LinkedListIntersect
{
    public static void Main(String[] args)
    {
        // list 1
        Node n1 = new Node(1);
        n1.next = new Node(2);
        n1.next.next = new Node(3);
        n1.next.next.next = new Node(4);
        n1.next.next.next.next = new Node(5);
        n1.next.next.next.next.next = new Node(6);
        n1.next.next.next.next.next.next = new Node(7);
        // list 2
        Node n2 = new Node(10);
        n2.next = new Node(9);
        n2.next.next = new Node(8);
        n2.next.next.next = n1.next.next.next;
        Print(n1);
        Print(n2);
        Console.WriteLine(MegeNode(n1, n2).data);
    }
 
    // function to print the list
    public static void Print(Node n)
    {
        Node cur = n;
        while (cur != null)
        {
            Console.Write(cur.data + " ");
            cur = cur.next;
        }
        Console.WriteLine();
    }
 
    // function to find the intersection of two node
    public static Node MegeNode(Node n1, Node n2)
    {
        // define hashset
        HashSet<Node> hs = new HashSet<Node>();
        while (n1 != null)
        {
            hs.Add(n1);
            n1 = n1.next;
        }
        while (n2 != null)
        {
            if (hs.Contains(n2))
            {
                return n2;
            }
            n2 = n2.next;
        }
        return null;
    }
}
 
// This code is contributed by 29AjayKumar


Javascript




<script>
 
// Javascript program to get intersection
// point of two linked list
class Node
{
    constructor(d)
    {
        this.data = d;
        this.next = null;
    }
}
 
// Function to print the list
function Print(n)
{
    let cur = n;
    while (cur != null)
    {
        document.write(cur.data + "  ");
        cur = cur.next;
    }
    document.write("<br>");
}
 
// Function to find the intersection of two node
function MegeNode(n1, n2)
{
     
    // Define hashset
    let hs = new Set();
    while (n1 != null)
    {
        hs.add(n1);
        n1 = n1.next;
    }
    while (n2 != null)
    {
        if (hs.has(n2))
        {
            return n2;
        }
        n2 = n2.next;
    }
    return null;
}
 
// Driver code
 
// list 1
let n1 = new Node(1);
n1.next = new Node(2);
n1.next.next = new Node(3);
n1.next.next.next = new Node(4);
n1.next.next.next.next = new Node(5);
n1.next.next.next.next.next = new Node(6);
n1.next.next.next.next.next.next = new Node(7);
 
// list 2
let n2 = new Node(10);
n2.next = new Node(9);
n2.next.next = new Node(8);
n2.next.next.next = n1.next.next.next;
 
Print(n1);
Print(n2);
 
document.write(MegeNode(n1, n2).data);
 
// This code is contributed by rag2127
 
</script>


Output

1 2 3 4 5 6 7 
10 9 8 4 5 6 7 
4

This method required O(n) additional space and is not very efficient if one list is large.

Method 8( 2-pointer technique ):

Using Two pointers : 

  • Initialize two pointers ptr1 and ptr2  at head1 and  head2.
  • Traverse through the lists, one node at a time.
  • When ptr1 reaches the end of a list, then redirect it to head2.
  • similarly, when ptr2 reaches the end of a list, redirect it to the head1.
  • Once both of them go through reassigning, they will be equidistant from 
     the collision point
  • If at any node ptr1 meets ptr2, then it is the intersection node.
  • After the second iteration if there is no intersection node it returns NULL.

C++




// CPP program to print intersection of lists
#include <bits/stdc++.h>
using namespace std;
 
/* Link list node */
class Node {
public:
    int data;
    Node* next;
};
 
// A utility function to return  intersection node
Node* intersectPoint(Node* head1, Node* head2)
{
    // Maintaining two pointers ptr1 and ptr2
    // at the head of A and B,
    Node* ptr1 = head1;
    Node* ptr2 = head2;
 
    // If any one of head is NULL i.e
    // no Intersection Point
    if (ptr1 == NULL || ptr2 == NULL)
        return NULL;
 
    // Traverse through the lists until they
    // reach Intersection node
    while (ptr1 != ptr2) {
 
        ptr1 = ptr1->next;
        ptr2 = ptr2->next;
 
        // If at any node ptr1 meets ptr2, then it is
        // intersection node.Return intersection node.
 
        if (ptr1 == ptr2)
            return ptr1;
        /* Once both of them go through reassigning,
        they will be equidistant from the collision point.*/
 
        // When ptr1 reaches the end of a list, then
        // reassign it to the head2.
        if (ptr1 == NULL)
            ptr1 = head2;
        // When ptr2 reaches the end of a list, then
        // redirect it to the head1.
        if (ptr2 == NULL)
            ptr2 = head1;
    }
    return ptr1;
}
 
// Function to print intersection nodes
// in  a given linked list
void print(Node* node)
{
    if (node == NULL)
        cout << "NULL";
    while (node->next != NULL) {
        cout << node->data << "->";
        node = node->next;
    }
    cout << node->data;
}
// Driver code
int main()
{
    /*
    Create two linked lists
 
    1st Linked list is 3->6->9->15->30
    2nd Linked list is 10->15->30
 
    15 30 are elements in the intersection list
    */
 
    Node* newNode;
    Node* head1 = new Node();
    head1->data = 10;
    Node* head2 = new Node();
    head2->data = 3;
    newNode = new Node();
    newNode->data = 6;
    head2->next = newNode;
    newNode = new Node();
    newNode->data = 9;
    head2->next->next = newNode;
    newNode = new Node();
    newNode->data = 15;
    head1->next = newNode;
    head2->next->next->next = newNode;
    newNode = new Node();
    newNode->data = 30;
    head1->next->next = newNode;
    head1->next->next->next = NULL;
    Node* intersect_node = NULL;
 
    // Find the intersection node of two linked lists
    intersect_node = intersectPoint(head1, head2);
    cout << "INTERSEPOINT LIST :";
    print(intersect_node);
    return 0;
}
 
// This code is contributed by Aditya Kumar (adityakumar129)


C




// c program to print intersection of lists
#include <stdio.h>
#include <stdlib.h>
 
/* Link list node */
typedef struct Node {
    int data;
    struct Node* next;
}Node;
 
// A utility function to return  intersection node
Node* intersectPoint(Node* head1, Node* head2)
{
    // Maintaining two pointers ptr1 and ptr2
    // at the head of A and B,
    Node* ptr1 = head1;
    Node* ptr2 = head2;
 
    // If any one of head is NULL i.e
    // no Intersection Point
    if (ptr1 == NULL || ptr2 == NULL)
        return NULL;
 
    // Traverse through the lists until they
    // reach Intersection node
    while (ptr1 != ptr2) {
 
        ptr1 = ptr1->next;
        ptr2 = ptr2->next;
 
        // If at any node ptr1 meets ptr2, then it is
        // intersection node.Return intersection node.
 
        if (ptr1 == ptr2)
            return ptr1;
        /* Once both of them go through reassigning,
        they will be equidistant from the collision point.*/
 
        // When ptr1 reaches the end of a list, then
        // reassign it to the head2.
        if (ptr1 == NULL)
            ptr1 = head2;
        // When ptr2 reaches the end of a list, then
        // redirect it to the head1.
        if (ptr2 == NULL)
            ptr2 = head1;
    }
    return ptr1;
}
 
// Function to print intersection nodes
// in  a given linked list
void print(Node* node)
{
    if (node == NULL)
        printf("NULL");
    while (node->next != NULL) {
        printf("%d->", node->data);
        node = node->next;
    }
    printf("%d", node->data);
}
// Driver code
int main()
{
    /*
    Create two linked lists
 
    1st Linked list is 3->6->9->15->30
    2nd Linked list is 10->15->30
 
    15 30 are elements in the intersection list
    */
    Node* newNode;
    Node* head1 = (Node*)malloc(sizeof(Node));
    head1->data = 10;
    Node* head2 = (Node*)malloc(sizeof(Node));
    head2->data = 3;
    newNode = (Node*)malloc(sizeof(Node));
    newNode->data = 6;
    head2->next = newNode;
    newNode = (Node*)malloc(sizeof(Node));
    newNode->data = 9;
    head2->next->next = newNode;
    newNode = (Node*)malloc(sizeof(Node));
    newNode->data = 15;
    head1->next = newNode;
    head2->next->next->next = newNode;
    newNode = (Node*)malloc(sizeof(Node));
    newNode->data = 30;
    head1->next->next = newNode;
    head1->next->next->next = NULL;
    Node* intersect_node = NULL;
 
    // Find the intersection node of two linked lists
    intersect_node = intersectPoint(head1, head2);
    printf("INTERSEPOINT LIST :");
    print(intersect_node);
    return 0;
}
 
// This code is contributed by Aditya Kumar (adityakumar129)


Java




// JAVA program to print intersection of lists
 
import java.util.*;
 
class GFG{
 
/* Link list node */
static class Node {
    int data;
    Node next;
};
 
// A utility function to return  intersection node
static Node intersectPoint(Node head1, Node head2)
{
    // Maintaining two pointers ptr1 and ptr2
    // at the head of A and B,
    Node ptr1 = head1;
    Node ptr2 = head2;
 
    // If any one of head is null i.e
    // no Intersection Point
    if (ptr1 == null || ptr2 == null) {
 
        return null;
    }
 
    // Traverse through the lists until they
    // reach Intersection node
    while (ptr1 != ptr2) {
 
        ptr1 = ptr1.next;
        ptr2 = ptr2.next;
 
        // If at any node ptr1 meets ptr2, then it is
        // intersection node.Return intersection node.
 
        if (ptr1 == ptr2) {
 
            return ptr1;
        }
        /* Once both of them go through reassigning,
        they will be equidistant from the collision point.*/
 
        // When ptr1 reaches the end of a list, then
        // reassign it to the head2.
        if (ptr1 == null) {
 
            ptr1 = head2;
        }
        // When ptr2 reaches the end of a list, then
        // redirect it to the head1.
        if (ptr2 == null) {
 
            ptr2 = head1;
        }
    }
 
    return ptr1;
}
 
// Function to print intersection nodes
// in  a given linked list
static void print(Node node)
{
    if (node == null)
        System.out.print("null");
    while (node.next != null) {
        System.out.print(node.data+ ".");
        node = node.next;
    }
    System.out.print(node.data);
}
   
// Driver code
public static void main(String[] args)
{
    /*
    Create two linked lists
 
    1st Linked list is 3.6.9.15.30
    2nd Linked list is 10.15.30
 
    15 30 are elements in the intersection list
    */
 
    Node newNode;
    Node head1 = new Node();
    head1.data = 10;
    Node head2 = new Node();
    head2.data = 3;
    newNode = new Node();
    newNode.data = 6;
    head2.next = newNode;
    newNode = new Node();
    newNode.data = 9;
    head2.next.next = newNode;
    newNode = new Node();
    newNode.data = 15;
    head1.next = newNode;
    head2.next.next.next = newNode;
    newNode = new Node();
    newNode.data = 30;
    head1.next.next = newNode;
    head1.next.next.next = null;
    Node intersect_node = null;
 
    // Find the intersection node of two linked lists
    intersect_node = intersectPoint(head1, head2);
 
    System.out.print("INTERSEPOINT LIST :");
 
    print(intersect_node);
}
}
 
// This code is contributed by umadevi9616.


Python3




# Python3 program to print intersection of lists
 
#  Link list node
class Node:
    def __init__(self, data = 0, next = None):
        self.data = data
        self.next = next
 
# A utility function to return  intersection node
def intersectPoint(head1, head2):
 
    # Maintaining two pointers ptr1 and ptr2
    # at the head of A and B,
    ptr1 = head1
    ptr2 = head2
 
    # If any one of head is None i.e
    # no Intersection Point
    if (ptr1 == None or ptr2 == None):
        return None
 
    # Traverse through the lists until they
    # reach Intersection node
    while (ptr1 != ptr2):
 
        ptr1 = ptr1.next
        ptr2 = ptr2.next
 
      # If at any node ptr1 meets ptr2, then it is
      # intersection node.Return intersection node.
        if (ptr1 == ptr2):
            return ptr1
 
        # Once both of them go through reassigning,
        # they will be equidistant from the collision point.
 
        # When ptr1 reaches the end of a list, then
        # reassign it to the head2.
        if (ptr1 == None):
            ptr1 = head2
 
        # When ptr2 reaches the end of a list, then
        # redirect it to the head1.
        if (ptr2 == None):
            ptr2 = head1
 
    return ptr1
 
# Function to print intersection nodes
# in  a given linked list
def Print(node):
 
    if (node == None):
        print("None")
    while (node.next != None):
        print(node.data,end="->")
        node = node.next
    print(node.data)
 
# Driver code
 
# Create two linked lists
 
# 1st Linked list is 3->6->9->15->30
# 2nd Linked list is 10->15->30
 
# 15 30 are elements in the intersection list
 
head1 = Node()
head1.data = 10
head2 = Node()
head2.data = 3
newNode = Node()
newNode.data = 6
head2.next = newNode
newNode = Node()
newNode.data = 9
head2.next.next = newNode
newNode = Node()
newNode.data = 15
head1.next = newNode
head2.next.next.next = newNode
newNode = Node()
newNode.data = 30
head1.next.next = newNode
head1.next.next.next = None
intersect_node = None
 
  # Find the intersection node of two linked lists
intersect_node = intersectPoint(head1, head2)
 
print("INTERSEPOINT LIST :",end="")
 
Print(intersect_node)
 
# This code is contributed by shinjanpatra


C#




// C# program to print intersection of lists
using System;
public class GFG {
 
    /* Link list node */
    public
 
 class Node {
        public
 
 int data;
        public
 
 Node next;
    };
 
    // A utility function to return intersection node
    static Node intersectPoint(Node head1, Node head2) {
       
        // Maintaining two pointers ptr1 and ptr2
        // at the head of A and B,
        Node ptr1 = head1;
        Node ptr2 = head2;
 
        // If any one of head is null i.e
        // no Intersection Point
        if (ptr1 == null || ptr2 == null) {
 
            return null;
        }
 
        // Traverse through the lists until they
        // reach Intersection node
        while (ptr1 != ptr2) {
 
            ptr1 = ptr1.next;
            ptr2 = ptr2.next;
 
            // If at any node ptr1 meets ptr2, then it is
            // intersection node.Return intersection node.
 
            if (ptr1 == ptr2) {
 
                return ptr1;
            }
            /*
             * Once both of them go through reassigning, they will be equidistant from the
             * collision point.
             */
 
            // When ptr1 reaches the end of a list, then
            // reassign it to the head2.
            if (ptr1 == null) {
 
                ptr1 = head2;
            }
            // When ptr2 reaches the end of a list, then
            // redirect it to the head1.
            if (ptr2 == null) {
 
                ptr2 = head1;
            }
        }
 
        return ptr1;
    }
 
    // Function to print intersection nodes
    // in a given linked list
    static void print(Node node) {
        if (node == null)
            Console.Write("null");
        while (node.next != null) {
            Console.Write(node.data + "->");
            node = node.next;
        }
        Console.Write(node.data);
    }
 
    // Driver code
    public static void Main(String[] args)
    {
       
        /*
         * Create two linked lists
         *
         * 1st Linked list is 3.6.9.15.30 2nd Linked list is 10.15.30
         *
         * 15 30 are elements in the intersection list
         */
 
        Node newNode;
        Node head1 = new Node();
        head1.data = 10;
        Node head2 = new Node();
        head2.data = 3;
        newNode = new Node();
        newNode.data = 6;
        head2.next = newNode;
        newNode = new Node();
        newNode.data = 9;
        head2.next.next = newNode;
        newNode = new Node();
        newNode.data = 15;
        head1.next = newNode;
        head2.next.next.next = newNode;
        newNode = new Node();
        newNode.data = 30;
        head1.next.next = newNode;
        head1.next.next.next = null;
        Node intersect_node = null;
 
        // Find the intersection node of two linked lists
        intersect_node = intersectPoint(head1, head2);
 
        Console.Write("INTERSEPOINT LIST :");
 
        print(intersect_node);
    }
}
 
// This code is contributed by gauravrajput1


Javascript




<script>
// Javascript program to print intersection of lists
 
 
 
 
/* Link list node */
class Node {
    constructor() {
        this.data = null;
        this.next = null;
    }
};
 
// A utility function to return  intersection node
function intersectPoint(head1, head2) {
    // Maintaining two pointers ptr1 and ptr2
    // at the head of A and B,
    let ptr1 = head1;
    let ptr2 = head2;
 
    // If any one of head is null i.e
    // no Intersection Point
    if (ptr1 == null || ptr2 == null) {
 
        return null;
    }
 
    // Traverse through the lists until they
    // reach Intersection node
    while (ptr1 != ptr2) {
 
        ptr1 = ptr1.next;
        ptr2 = ptr2.next;
 
        // If at any node ptr1 meets ptr2, then it is
        // intersection node.Return intersection node.
 
        if (ptr1 == ptr2) {
 
            return ptr1;
        }
        /* Once both of them go through reassigning,
        they will be equidistant from the collision point.*/
 
        // When ptr1 reaches the end of a list, then
        // reassign it to the head2.
        if (ptr1 == null) {
 
            ptr1 = head2;
        }
        // When ptr2 reaches the end of a list, then
        // redirect it to the head1.
        if (ptr2 == null) {
 
            ptr2 = head1;
        }
    }
 
    return ptr1;
}
 
// Function to print intersection nodes
// in  a given linked list
function print(node) {
    if (node == null)
        document.write("null");
    while (node.next != null) {
        document.write(node.data + "->");
        node = node.next;
    }
    document.write(node.data);
}
 
// Driver code
/*
Create two linked lists
 
1st Linked list is 3.6.9.15.30
2nd Linked list is 10.15.30
 
15 30 are elements in the intersection list
*/
 
let newNode;
let head1 = new Node();
head1.data = 10;
let head2 = new Node();
head2.data = 3;
newNode = new Node();
newNode.data = 6;
head2.next = newNode;
newNode = new Node();
newNode.data = 9;
head2.next.next = newNode;
newNode = new Node();
newNode.data = 15;
head1.next = newNode;
head2.next.next.next = newNode;
newNode = new Node();
newNode.data = 30;
head1.next.next = newNode;
head1.next.next.next = null;
let intersect_node = null;
 
// Find the intersection node of two linked lists
intersect_node = intersectPoint(head1, head2);
 
document.write("INTERSEPOINT LIST :");
 
print(intersect_node);
 
// This code is contributed by Saurabh Jaiswal
</script>


Output

INTERSEPOINT LIST :15->30

Time complexity : O( m + n ) 
Auxiliary Space:  O(1)

Please write comments if you find any bug in the above algorithm or a better way to solve the same problem.


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