Merge a linked list into another linked list at alternate positions
Given two linked lists, insert nodes of second list into first list at alternate positions of first list.
For example, if first list is 5->7->17->13->11 and second is 12->10->2->4->6, the first list should become 5->12->7->10->17->2->13->4->11->6 and second list should become empty. The nodes of second list should only be inserted when there are positions available. For example, if the first list is 1->2->3 and second list is 4->5->6->7->8, then first list should become 1->4->2->5->3->6 and second list to 7->8.
Use of extra space is not allowed (Not allowed to create additional nodes), i.e., insertion must be done in-place. Expected time complexity is O(n) where n is number of nodes in first list.
The idea is to run a loop while there are available positions in first loop and insert nodes of second list by changing pointers.
Following are implementations of this approach.
C++
// C++ program to merge a linked list into another at // alternate positions #include <bits/stdc++.h> using namespace std; // A nexted list node class Node { public : int data; Node *next; }; /* Function to insert a node at the beginning */ void push(Node ** head_ref, int new_data) { Node* new_node = new Node(); new_node->data = new_data; new_node->next = (*head_ref); (*head_ref) = new_node; } /* Utility function to print a singly linked list */ void printList(Node *head) { Node *temp = head; while (temp != NULL) { cout<<temp->data<< " " ; temp = temp->next; } cout<<endl; } // Main function that inserts nodes of linked list q into p at // alternate positions. Since head of first list never changes // and head of second list may change, we need single pointer // for first list and double pointer for second list. void merge(Node *p, Node **q) { Node *p_curr = p, *q_curr = *q; Node *p_next, *q_next; // While there are available positions in p while (p_curr != NULL && q_curr != NULL) { // Save next pointers p_next = p_curr->next; q_next = q_curr->next; // Make q_curr as next of p_curr q_curr->next = p_next; // Change next pointer of q_curr p_curr->next = q_curr; // Change next pointer of p_curr // Update current pointers for next iteration p_curr = p_next; q_curr = q_next; } *q = q_curr; // Update head pointer of second list } // Driver code int main() { Node *p = NULL, *q = NULL; push(&p, 3); push(&p, 2); push(&p, 1); cout<< "First Linked List:\n" ; printList(p); push(&q, 8); push(&q, 7); push(&q, 6); push(&q, 5); push(&q, 4); cout<< "Second Linked List:\n" ; printList(q); merge(p, &q); cout<< "Modified First Linked List:\n" ; printList(p); cout<< "Modified Second Linked List:\n" ; printList(q); return 0; } // This code is contributed by rathbhupendra |
C
// C program to merge a linked list into another at // alternate positions #include <stdio.h> #include <stdlib.h> // A nexted list node struct Node { int data; struct Node *next; }; /* Function to insert a node at the beginning */ void push( struct Node ** head_ref, int new_data) { struct Node* new_node = ( struct Node*) malloc ( sizeof ( struct Node)); new_node->data = new_data; new_node->next = (*head_ref); (*head_ref) = new_node; } /* Utility function to print a singly linked list */ void printList( struct Node *head) { struct Node *temp = head; while (temp != NULL) { printf ( "%d " , temp->data); temp = temp->next; } printf ( "\n" ); } // Main function that inserts nodes of linked list q into p at // alternate positions. Since head of first list never changes // and head of second list may change, we need single pointer // for first list and double pointer for second list. void merge( struct Node *p, struct Node **q) { struct Node *p_curr = p, *q_curr = *q; struct Node *p_next, *q_next; // While there are available positions in p while (p_curr != NULL && q_curr != NULL) { // Save next pointers p_next = p_curr->next; q_next = q_curr->next; // Make q_curr as next of p_curr q_curr->next = p_next; // Change next pointer of q_curr p_curr->next = q_curr; // Change next pointer of p_curr // Update current pointers for next iteration p_curr = p_next; q_curr = q_next; } *q = q_curr; // Update head pointer of second list } // Driver program to test above functions int main() { struct Node *p = NULL, *q = NULL; push(&p, 3); push(&p, 2); push(&p, 1); printf ( "First Linked List:\n" ); printList(p); push(&q, 8); push(&q, 7); push(&q, 6); push(&q, 5); push(&q, 4); printf ( "Second Linked List:\n" ); printList(q); merge(p, &q); printf ( "Modified First Linked List:\n" ); printList(p); printf ( "Modified Second Linked List:\n" ); printList(q); getchar (); return 0; } |
Java
// Java program to merge a linked list into another at // alternate positions class LinkedList { Node head; // head of list /* Linked list Node*/ class Node { int data; Node next; Node( int d) {data = d; next = null ; } } /* Inserts a new Node at front of the list. */ void push( int new_data) { /* 1 & 2: Allocate the Node & Put in the data*/ Node new_node = new Node(new_data); /* 3. Make next of new Node as head */ new_node.next = head; /* 4. Move the head to point to new Node */ head = new_node; } // Main function that inserts nodes of linked list q into p at // alternate positions. Since head of first list never changes // and head of second list/ may change, we need single pointer // for first list and double pointer for second list. void merge(LinkedList q) { Node p_curr = head, q_curr = q.head; Node p_next, q_next; // While there are available positions in p; while (p_curr != null && q_curr != null ) { // Save next pointers p_next = p_curr.next; q_next = q_curr.next; // make q_curr as next of p_curr q_curr.next = p_next; // change next pointer of q_curr p_curr.next = q_curr; // change next pointer of p_curr // update current pointers for next iteration p_curr = p_next; q_curr = q_next; } q.head = q_curr; } /* Function to print linked list */ void printList() { Node temp = head; while (temp != null ) { System.out.print(temp.data+ " " ); temp = temp.next; } System.out.println(); } /* Driver program to test above functions */ public static void main(String args[]) { LinkedList llist1 = new LinkedList(); LinkedList llist2 = new LinkedList(); llist1.push( 3 ); llist1.push( 2 ); llist1.push( 1 ); System.out.println( "First Linked List:" ); llist1.printList(); llist2.push( 8 ); llist2.push( 7 ); llist2.push( 6 ); llist2.push( 5 ); llist2.push( 4 ); System.out.println( "Second Linked List:" ); llist1.merge(llist2); System.out.println( "Modified first linked list:" ); llist1.printList(); System.out.println( "Modified second linked list:" ); llist2.printList(); } } /* This code is contributed by Rajat Mishra */ |
Python3
# Python program to merge a linked list into another at # alternate positions class Node( object ): def __init__( self , data: int ): self .data = data self . next = None class LinkedList( object ): def __init__( self ): self .head = None def push( self , new_data: int ): new_node = Node(new_data) new_node. next = self .head # 4. Move the head to point to new Node self .head = new_node # Function to print linked list from the Head def printList( self ): temp = self .head while temp ! = None : print (temp.data) temp = temp. next # Main function that inserts nodes of linked list q into p at alternate positions. # Since head of first list never changes # but head of second list/ may change, # we need single pointer for first list and double pointer for second list. def merge( self , p, q): p_curr = p.head q_curr = q.head # swap their positions until one finishes off while p_curr ! = None and q_curr ! = None : # Save next pointers p_next = p_curr. next q_next = q_curr. next # make q_curr as next of p_curr q_curr. next = p_next # change next pointer of q_curr p_curr. next = q_curr # change next pointer of p_curr # update current pointers for next iteration p_curr = p_next q_curr = q_next q.head = q_curr # Driver program to test above functions llist1 = LinkedList() llist2 = LinkedList() # Creating LLs # 1. llist1.push( 3 ) llist1.push( 2 ) llist1.push( 1 ) llist1.push( 0 ) # 2. for i in range ( 8 , 3 , - 1 ): llist2.push(i) print ( "First Linked List:" ) llist1.printList() print ( "Second Linked List:" ) llist2.printList() # Merging the LLs llist1.merge(p = llist1, q = llist2) print ( "Modified first linked list:" ) llist1.printList() print ( "Modified second linked list:" ) llist2.printList() # This code is contributed by Deepanshu Mehta |
C#
// C# program to merge a linked list into // another at alternate positions using System; public class LinkedList { Node head; // head of list /* Linked list Node*/ public class Node { public int data; public Node next; public Node( int d) { data = d; next = null ; } } /* Inserts a new Node at front of the list. */ void push( int new_data) { /* 1 & 2: Allocate the Node & Put in the data*/ Node new_node = new Node(new_data); /* 3. Make next of new Node as head */ new_node.next = head; /* 4. Move the head to point to new Node */ head = new_node; } // Main function that inserts nodes // of linked list q into p at alternate // positions. Since head of first list // never changes and head of second // list/ may change, we need single // pointer for first list and double // pointer for second list. void merge(LinkedList q) { Node p_curr = head, q_curr = q.head; Node p_next, q_next; // While there are available positions in p; while (p_curr != null && q_curr != null ) { // Save next pointers p_next = p_curr.next; q_next = q_curr.next; // make q_curr as next of p_curr q_curr.next = p_next; // change next pointer of q_curr p_curr.next = q_curr; // change next pointer of p_curr // update current pointers for next iteration p_curr = p_next; q_curr = q_next; } q.head = q_curr; } /* Function to print linked list */ void printList() { Node temp = head; while (temp != null ) { Console.Write(temp.data+ " " ); temp = temp.next; } Console.WriteLine(); } /* Driver code*/ public static void Main() { LinkedList llist1 = new LinkedList(); LinkedList llist2 = new LinkedList(); llist1.push(3); llist1.push(2); llist1.push(1); Console.WriteLine( "First Linked List:" ); llist1.printList(); llist2.push(8); llist2.push(7); llist2.push(6); llist2.push(5); llist2.push(4); Console.WriteLine( "Second Linked List:" ); llist1.merge(llist2); Console.WriteLine( "Modified first linked list:" ); llist1.printList(); Console.WriteLine( "Modified second linked list:" ); llist2.printList(); } } /* This code contributed by PrinciRaj1992 */ |
Javascript
<script> // Javascript program to merge a linked list into another at // alternate positions // A nexted list node class Node { constructor() { this .data = 0; this .next = null ; } }; /* Function to insert a node at the beginning */ function push(head_ref, new_data) { var new_node = new Node(); new_node.data = new_data; new_node.next = (head_ref); (head_ref) = new_node; return head_ref; } /* Utility function to print a singly linked list */ function printList(head) { var temp = head; while (temp != null ) { document.write( temp.data + " " ); temp = temp.next; } document.write( "<br>" ); } // Main function that inserts nodes of linked list q into p at // alternate positions. Since head of first list never changes // and head of second list may change, we need single pointer // for first list and double pointer for second list. function merge(p, q) { var p_curr = p, q_curr = q; var p_next, q_next; // While there are available positions in p while (p_curr != null && q_curr != null ) { // Save next pointers p_next = p_curr.next; q_next = q_curr.next; // Make q_curr as next of p_curr q_curr.next = p_next; // Change next pointer of q_curr p_curr.next = q_curr; // Change next pointer of p_curr // Update current pointers for next iteration p_curr = p_next; q_curr = q_next; } q = q_curr; // Update head pointer of second list return q; } // Driver code var p = null , q = null ; p = push(p, 3); p = push(p, 2); p = push(p, 1); document.write( "First Linked List:<br>" ); printList(p); q = push(q, 8); q = push(q, 7); q = push(q, 6); q = push(q, 5); q = push(q, 4); document.write( "Second Linked List:<br>" ); printList(q); q = merge(p, q); document.write( "Modified First Linked List:<br>" ); printList(p); document.write( "Modified Second Linked List:<br>" ); printList(q); // This code is contributed by rrrtnx. </script> |
First Linked List: 1 2 3 Second Linked List: 4 5 6 7 8 Modified First Linked List: 1 4 2 5 3 6 Modified Second Linked List: 7 8
Time Complexity: O(min(n1, n2)), where n1 and n2 represents the length of the given two linked lists.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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