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Merge a linked list into another linked list at alternate positions

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  • Difficulty Level : Easy
  • Last Updated : 03 Jul, 2022
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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. 

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The idea is to run a loop while there are available positions in first loop and insert nodes of second list by changing pointers.

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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>


Output

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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