Find intersection point of two Linked Lists without finding the length
Last Updated :
10 Apr, 2023
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 both the linked lists merge.
Examples:
Input: 1 -> 2 -> 3 -> 4 -> 5 -> 6
^
|
7 -> 8 -> 9
Output: 4
Input: 13 -> 14 -> 5 -> 6
^
|
10 -> 2 -> 3 -> 4
Output: 14
Prerequisites: Write a function to get the intersection point of two Linked Lists
Approach: Take two pointers for the heads of both the linked lists. If one of them reaches the end earlier then use it by moving it to the beginning of the other list. Once both of them go through reassigning they will be equidistance from the collision point.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
/* Link list node */
class Node {
public:
int data;
Node* next;
};
// Function to return the intersection point
// of the two linked lists head1 and head2
int getIntesectionNode(Node* head1, Node* head2)
{
Node* current1 = head1;
Node* current2 = head2;
// If one of the head is NULL
if (!current1 or !current2)
return -1;
// Continue until we find intersection node
while (current1 and current2
and current1 != current2) {
current1 = current1->next;
current2 = current2->next;
// If we get intersection node
if (current1 == current2)
return current1->data;
// If one of them reaches end
if (!current1)
current1 = head2;
if (!current2)
current2 = head1;
}
return current1->data;
}
// 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 << getIntesectionNode(head1, head2);
return 0;
}
// Java implementation of the approach
class GFG
{
/* Link list node */
static class Node
{
int data;
Node next;
};
// Function to return the intersection point
// of the two linked lists head1 and head2
static int getIntesectionNode(Node head1, Node head2)
{
Node current1 = head1;
Node current2 = head2;
// If one of the head is null
if (current1 == null || current2 == null )
return -1;
// Continue until we find intersection node
while (current1 != null && current2 != null
&& current1 != current2)
{
current1 = current1.next;
current2 = current2.next;
// If we get intersection node
if (current1 == current2)
return current1.data;
// If one of them reaches end
if (current1 == null )
current1 = head2;
if (current2 == null )
current2 = head1;
}
return current1.data;
}
// Driver code
public static void main(String[] args)
{
/*
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;
System.out.print(getIntesectionNode(head1, head2));
}
}
// This code is contributed by 29AjayKumar
# Python3 implementation of the approach
''' Link list node '''
class new_Node:
# Constructor to initialize the node object
def __init__(self, data):
self.data = data
self.next = None
# Function to return the intersection point
# of the two linked lists head1 and head2
def getIntesectionNode(head1, head2):
current1 = head1
current2 = head2
# If one of the head is None
if (not current1 or not current2 ):
return -1
# Continue until we find intersection node
while (current1 and current2 and current1 != current2):
current1 = current1.next
current2 = current2.next
# If we get intersection node
if (current1 == current2):
return current1.data
# If one of them reaches end
if (not current1):
current1 = head2
if (not current2):
current2 = head1
return current1.data
# Driver code
'''
Create two linked lists
1st 3.6.9.15.30
2nd 10.15.30
15 is the intersection po
'''
# Addition of newNodes
head1 = new_Node(10)
head2 = new_Node(3)
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 = None
print(getIntesectionNode(head1, head2))
# This code is contributed by shubhamsingh10
// C# implementation of the approach
using System;
class GFG
{
/* Link list node */
class Node
{
public int data;
public Node next;
};
// Function to return the intersection point
// of the two linked lists head1 and head2
static int getIntesectionNode(Node head1, Node head2)
{
Node current1 = head1;
Node current2 = head2;
// If one of the head is null
if (current1 == null || current2 == null )
return -1;
// Continue until we find intersection node
while (current1 != null && current2 != null
&& current1 != current2)
{
current1 = current1.next;
current2 = current2.next;
// If we get intersection node
if (current1 == current2)
return current1.data;
// If one of them reaches end
if (current1 == null )
current1 = head2;
if (current2 == null )
current2 = head1;
}
return current1.data;
}
// Driver code
public static void Main(String[] args)
{
/*
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;
Console.Write(getIntesectionNode(head1, head2));
}
}
// This code is contributed by PrinciRaj1992
<script>
// Javascript implementation of the approach
/* Link list node */
class Node {
constructor()
{
this.data = 0;
this.next = null;
}
};
// Function to return the intersection point
// of the two linked lists head1 and head2
function getIntesectionNode(head1, head2)
{
var current1 = head1;
var current2 = head2;
// If one of the head is null
if (!current1 || !current2)
return -1;
// Continue until we find intersection node
while (current1 && current2
&& current1 != current2) {
current1 = current1.next;
current2 = current2.next;
// If we get intersection node
if (current1 == current2)
return current1.data;
// If one of them reaches end
if (!current1)
current1 = head2;
if (!current2)
current2 = head1;
}
return current1.data;
}
// Driver code
/*
Create two linked lists
1st 3.6.9.15.30
2nd 10.15.30
15 is the intersection point
*/
var newNode;
// Addition of new nodes
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;
document.write( getIntesectionNode(head1, head2));
// This code is contributed by noob2000.
</script>
Time complexity O(m+n),where m and n are the lengths of the two linked lists
Space complexity O(1),as it uses a constant amount of additional memory to store only a few pointers to nodes.
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