Get the Intersection point of two Linked List by counting nodes

Last Updated : 28 Nov, 2022

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.

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.
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.
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:

C++

// 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; 
} 

Java

// Java implementation of the approach 
class GFG 
{ 
  
// Link list node 
static class Node  
{ 
    int data; 
    Node next; 
}; 
  
// Function to get the intersection point 
// of the given linked lists 
static 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 
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 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 
    System.out.print(getIntersectionNode(head1, head2)); 
} 
} 
  
// This code is contributed by 29AjayKumar 

Python3

# Python3 implementation of the approach 
  
# Link list node 
class Node: 
    def __init__(self): 
        self.data = 0
        self.next = None
  
# Function to get the intersection point 
# of the given linked lists 
def getIntersectionNode( head1, head2): 
  
    curr1 = head1 
    curr2 = head2 
  
    # While both the pointers are not equal 
    while (curr1 != curr2):  
  
        # If the first pointer is None then 
        # set it to point to the head of 
        # the second linked list 
        if (curr1 == None) : 
            curr1 = head2 
          
        # Else point it to the next node 
        else:  
            curr1 = curr1.next
          
        # If the second pointer is None then 
        # set it to point to the head of 
        # the first linked list 
        if (curr2 == None):  
            curr2 = head1 
          
        # Else point it to the next node 
        else:  
            curr2 = curr2.next
          
    # Return the intersection node 
    return curr1.data 
  
# Driver code 
  
# 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 
  
newNode = None
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
  
# Print the intersection node 
print( getIntersectionNode(head1, head2)) 
  
# This code is contributed by Arnab Kundu 

C#

// C# implementation of the approach 
using System; 
  
class GFG 
{ 
  
// Link list node 
public class Node  
{ 
    public int data; 
    public Node next; 
}; 
  
// Function to get the intersection point 
// of the given linked lists 
static 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 
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 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 
    Console.Write(getIntersectionNode(head1, head2)); 
} 
} 
  
// This code is contributed by Rajput-Ji 

Javascript

<script> 
  
// Javascript implementation of the approach 
  
// Link list node 
  
class Node { 
        constructor(data) { 
        this.data = data; 
        this.next = null; 
            } 
} 
  
// Function to get the intersection point 
// of the given linked lists 
function getIntersectionNode(head1, head2) 
{ 
    var 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 
  
/* 
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 
*/
  
var newNode; 
var head1 = new Node(); 
head1.data = 10; 
var 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 
document.write(getIntersectionNode(head1, head2)); 
  
// This code is contributed by jana_sayantan. 
</script>

Output:

Time Complexity: O(N + M).
Auxiliary Space: O(1).

C++ Program For Finding Intersection Point Of Two Linked Lists

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Get the Intersection point of two Linked List by counting nodes

C++

Java

Python3

C#

Javascript

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