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.
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 = Nonehead1 = Node() head1.data = 10head2 = Node() head2.data = 3newNode = Node() newNode.data = 6head2.next = newNode newNode = Node() newNode.data = 9head2.next.next = newNode newNode = Node() newNode.data = 15head1.next = newNode head2.next.next.next = newNode newNode = Node() newNode.data = 30head1.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:
15
Time Complexity: O(N + M).
Auxiliary Space: O(1).
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