Find intersection point of two Linked Lists without finding the length
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 head2int 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 codeint 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
// Java implementation of the approachclass GFG{/* Link list node */static class Node{ int data; Node next;};// Function to return the intersection point// of the two linked lists head1 and head2static 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 codepublic 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
# 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 head2def 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 newNodeshead1 = new_Node(10)head2 = new_Node(3)newNode = new_Node(6)head2.next = newNodenewNode = new_Node(9)head2.next.next = newNodenewNode = new_Node(15)head1.next = newNodehead2.next.next.next = newNodenewNode = new_Node(30)head1.next.next = newNodehead1.next.next.next = Noneprint(getIntesectionNode(head1, head2))# This code is contributed by shubhamsingh10 |
C#
// C# implementation of the approachusing 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 head2static 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 codepublic 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 |
Javascript
<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 head2function 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 nodesvar 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> |
Output:
15
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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