Write a function to get the intersection point of two Linked Lists | Set 2

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

Y ShapedLinked List
Above diagram shows an example with two linked list having 15 as intersection point.



Approach: It can be observed that the number of nodes in traversing the first linked list and then from the head of the second linked list to intersection point is equal to the number of nodes involved in traversing the second linked list and then from head of the first list to the intersection point. Considering the example given above, start traversing the two linked lists with two pointers curr1 and curr2 pointing to the heads of the given linked lists respectively.



  1. If curr1 != null then update it to point to the next node, else it is updated to point to the first node of the second list.
  2. If curr2 != null then update it to point to the next node, else it is updated to point to the first node of the first list.
  3. Repeat the above steps while curr1 is not equal to curr2.

The two pointers curr1 and curr2 will be pointing to the same node now i.e. the merging point.

Below is the implementation of the above approach:

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// C++ implementation of the approach
#include <iostream>
using namespace std;
  
// Link list node
struct Node {
    int data;
    Node* next;
};
  
// Function to get the intersection point
// of the given linked lists
int getIntersectionNode(Node* head1, Node* head2)
{
    Node *curr1 = head1, *curr2 = head2;
  
    // While both the pointers are not equal
    while (curr1 != curr2) {
  
        // If the first pointer is null then
        // set it to point to the head of
        // the second linked list
        if (curr1 == NULL) {
            curr1 = head2;
        }
  
        // Else point it to the next node
        else {
            curr1 = curr1->next;
        }
  
        // If the second pointer is null then
        // set it to point to the head of
        // the first linked list
        if (curr2 == NULL) {
            curr2 = head1;
        }
  
        // Else point it to the next node
        else {
            curr2 = curr2->next;
        }
    }
  
    // Return the intersection node
    return curr1->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 is the intersection point
    */
  
    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;
  
    // Print the intersection node
    cout << getIntersectionNode(head1, head2);
  
    return 0;
}

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Output:

15


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