C++ Program To Merge A Linked List Into Another Linked List At Alternate Positions

Last Updated : 16 Jun, 2022

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 
        // Change next pointer of q_curr 
        q_curr->next = p_next; 

        // Change next pointer of p_curr 
        p_curr->next = q_curr; 

        // Update current pointers for 
        // next iteration 
        p_curr = p_next; 
        q_curr = q_next; 
    } 

    // Update head pointer of second list 
    *q = q_curr; 
} 

// Driver code 
int main() 
{ 
    Node *p = NULL, *q = NULL; 
    push(&p, 3); 
    push(&p, 2); 
    push(&p, 1); 
    cout << "First Linked List:"; 
    printList(p); 

    push(&q, 8); 
    push(&q, 7); 
    push(&q, 6); 
    push(&q, 5); 
    push(&q, 4); 
    cout << "Second Linked List:"; 
    printList(q); 

    merge(p, &q); 

    cout << 
    "Modified First Linked List:"; 
    printList(p); 

    cout << 
    "Modified Second Linked List:"; 
    printList(q); 

    return 0; 
} 
// This code is contributed by rathbhupendra

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.

Please refer complete article on Merge a linked list into another linked list at alternate positions for more details!