Given two linked lists of size N and M consisting of positive value nodes, having a common intersection point, the task is to find the intersection point of the two linked lists where they merge.
Examples:
Input: L1: 3 ? 6 ? 9 ? 15 ? 30, L2: 10 ? 15 ? 30
Output: 15
Explanation:From the above image, the intersection point of the two linked lists is 15.
Input: L1: 1 ? 2 ? 3, L2: 4 ? 5 ? 1 ? 2 ? 3
Output: 1
Approach: The idea is to traverse the first linked list and multiply the value of each node by -1 thus making them negative. Then, traverse the second linked list and print the value of the first node having a negative value. Follow the steps below to solve the problem:
- Traverse the first linked list L1 and multiply the value of each node by -1.
- Now, traverse the second linked list L2 and if there exists any node with negative values then print the absolute value of the node’s value as the resultant intersection of the linked list and break out of the loop.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Structure of a node // of a Linked List class Node {
public:
int data;
Node* next;
// Constructor
Node(int x)
{
data = x;
next = NULL;
}
}; // Function to find the intersection // point of the two Linked Lists Node* intersectingNode(Node* headA, Node* headB)
{ // Traverse the first linked list
// and multiply all values by -1
Node* a = headA;
while (a) {
// Update a -> data
a->data *= -1;
// Update a
a = a->next;
}
// Traverse the second Linked List
// and find the value of the first
// node having negative value
Node* b = headB;
while (b) {
// Intersection point
if (b->data < 0)
break;
// Update b
b = b->next;
}
return b;
} // Function to create linked lists void formLinkList(Node*& head1,
Node*& head2)
{ // Linked List L1
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);
// Linked List L2
head2 = new Node(10);
head2->next = head1->next->next->next;
return;
} // Driver Code int main()
{ Node* head1;
Node* head2;
formLinkList(head1, head2);
cout << abs(intersectingNode(head1,
head2)
->data);
return 0;
} |
Java
// Java program for the above approach import java.io.*;
class GFG{
static Node head1 = null;
static Node head2 = null;
// Structure of a node // of a Linked List static class Node
{ int data;
Node next;
// Constructor
Node(int x)
{
data = x;
next = null;
}
} // Function to find the intersection // point of the two Linked Lists static Node intersectingNode(Node headA, Node headB)
{ // Traverse the first linked list
// and multiply all values by -1
Node a = headA;
while (a != null)
{
// Update a -> data
a.data *= -1;
// Update a
a = a.next;
}
// Traverse the second Linked List
// and find the value of the first
// node having negative value
Node b = headB;
while (b != null)
{
// Intersection point
if (b.data < 0)
break;
// Update b
b = b.next;
}
return b;
} // Function to create linked lists static void formLinkList()
{ // Linked List L1
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);
// Linked List L2
head2 = new Node(10);
head2.next = head1.next.next.next;
return;
} // Driver Code public static void main(String[] args)
{ formLinkList();
System.out.println(Math.abs(
intersectingNode(head1, head2).data));
} } // This code is contributed by Dharanendra L V. |
Python3
# Python3 program for the above approach # Structure of a node # of a Linked List class Node:
def __init__(self, d):
self.data = d
self.next = None
# Function to find the intersection # point of the two Linked Lists def intersectingNode(headA, headB):
# Traverse the first linked list
# and multiply all values by -1
a = headA
while (a):
# Update a . data
a.data *= -1
# Update a
a = a.next
# Traverse the second Linked List
# and find the value of the first
# node having negative value
b = headB
while (b):
# Intersection point
if (b.data < 0):
break
# Update b
b = b.next
return b
# Function to create linked lists def formLinkList(head1, head2):
# Linked List L1
head1 = Node(3)
head1.next = Node(6)
head1.next.next = Node(9)
head1.next.next.next = Node(15)
head1.next.next.next.next = Node(30)
# Linked List L2
head2 = Node(10)
head2.next = head1.next.next.next
return head1, head2
# Driver Code if __name__ == '__main__':
head1, head2 = formLinkList(None, None)
print(abs(intersectingNode(head1, head2).data))
# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approach using System;
public class Node
{ public int data;
public Node next;
// Constructor
public Node(int x)
{
data = x;
next = null;
}
} public class GFG{
static Node head1 = null;
static Node head2 = null;
// Function to find the intersection // point of the two Linked Lists static Node intersectingNode(Node headA, Node headB)
{ // Traverse the first linked list
// and multiply all values by -1
Node a = headA;
while (a != null)
{
// Update a -> data
a.data *= -1;
// Update a
a = a.next;
}
// Traverse the second Linked List
// and find the value of the first
// node having negative value
Node b = headB;
while (b != null)
{
// Intersection point
if (b.data < 0)
break;
// Update b
b = b.next;
}
return b;
} // Function to create linked lists static void formLinkList()
{ // Linked List L1
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);
// Linked List L2
head2 = new Node(10);
head2.next = head1.next.next.next;
return;
} // Driver Code static public void Main ()
{
formLinkList();
Console.WriteLine(Math.Abs(
intersectingNode(head1, head2).data));
}
} // This code is contributed by unknown2108. |
Javascript
<script> // JavaScript program for the above approach let head1 = null;
let head2 = null;
// Structure of a node // of a Linked List class Node { constructor(x)
{
this.data=x;
this.next=null;
}
} // Function to find the intersection // point of the two Linked Lists function intersectingNode(headA,headB)
{ // Traverse the first linked list
// and multiply all values by -1
let a = headA;
while (a != null)
{
// Update a -> data
a.data *= -1;
// Update a
a = a.next;
}
// Traverse the second Linked List
// and find the value of the first
// node having negative value
let b = headB;
while (b != null)
{
// Intersection point
if (b.data < 0)
break;
// Update b
b = b.next;
}
return b;
} // Function to create linked lists function formLinkList()
{ // Linked List L1
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);
// Linked List L2
head2 = new Node(10);
head2.next = head1.next.next.next;
return;
} // Driver Code formLinkList(); document.write(Math.abs( intersectingNode(head1, head2).data)); // This code is contributed by patel2127 </script> |
Output:
15
Time Complexity: O(N + M)
Auxiliary Space: O(1)