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C++ Program For In-Place Merge Two Linked Lists Without Changing Links Of First List

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Given two sorted singly linked lists having n and m elements each, merge them using constant space. First, n smallest elements in both the lists should become part of the first list and the rest elements should be part of the second list. Sorted order should be maintained. We are not allowed to change pointers of the first linked list.


First List: 2->4->7->8->10
Second List: 1->3->12

First List: 1->2->3->4->7
Second List: 8->10->12

We strongly recommend you to minimize your browser and try this yourself first.
The problem becomes very simple if we’re allowed to change pointers of the first linked list. If we are allowed to change links, we can simply do something like a merge of the merge-sort algorithm. We assign the first n smallest elements to the first linked list where n is the number of elements in the first linked list and the rest to the second linked list. We can achieve this in O(m + n) time and O(1) space, but this solution violates the requirement that we can’t change links of the first list.

The problem becomes a little tricky as we’re not allowed to change pointers in the first linked list. The idea is something similar to this post but as we are given a singly linked list, we can’t proceed backward with the last element of LL2. 

The idea is for each element of LL1, we compare it with the first element of LL2. If LL1 has a greater element than the first element of LL2, then we swap the two elements involved. To keep LL2 sorted, we need to place the first element of LL2 at its correct position. We can find a mismatch by traversing LL2 once and correcting the pointers. 

Below is the implementation of this idea. 


// C++ Program to merge two sorted linked lists
// without using any extra space and without
// changing links of first list
#include <bits/stdc++.h>
using namespace std;
// Structure for a linked list node
struct Node
    int data;
    struct Node *next;
/* Given a reference (pointer to pointer)
   to the head of a list and an int, push
   a new node on the front of the list. */
void push(struct Node** head_ref,
          int new_data)
    // Allocate node 
    struct Node* new_node =
           (struct Node*) malloc(sizeof(struct Node));
    // Put in the data 
    new_node->data = new_data;
    /* Link the old list off the
       new node */
    new_node->next = (*head_ref);
    /* Move the head to point to
       the new node */
    (*head_ref) = new_node;
// Function to merge two sorted
// linked lists LL1 and LL2 without
// using any extra space.
void mergeLists(struct Node *a,
                struct Node * &b)
    // Run till either one of a or b
    // runs out
    while (a && b)
        // For each element of LL1,
        // compare it with first element
        // of LL2.
        if (a->data > b->data)
            // Swap the two elements involved
            // if LL1 has a greater element
            swap(a->data, b->data);
            struct Node *temp = b;
            // To keep LL2 sorted, place
            // first element of LL2 at its
            // correct place
            if (b->next && b->data >
                b = b->next;
                struct Node *ptr= b,
                            *prev = NULL;
                // Find mismatch by traversing
                // the second linked list once
                while (ptr && ptr->data <
                    prev = ptr;
                    ptr = ptr -> next;
                // correct the pointers
                prev->next = temp;
                temp->next = ptr;
        // Move LL1 pointer to next element
        a = a->next;
// Code to print the linked link
void printList(struct Node *head)
    while (head)
        cout << head->data << "->" ;
        head = head->next;
    cout << "NULL" << endl;
// Driver code
int main()
    struct Node *a = NULL;
    push(&a, 10);
    push(&a, 8);
    push(&a, 7);
    push(&a, 4);
    push(&a, 2);
    struct Node *b = NULL;
    push(&b, 12);
    push(&b, 3);
    push(&b, 1);
    mergeLists(a, b);
    cout << "First List: ";
    cout << "Second List: ";
    return 0;

Output : 

First List: 1->2->3->4->7->NULL
Second List: 8->10->12->NULL

Time Complexity : O(mn)

Auxiliary Space: O(1)

Please refer complete article on In-place Merge two linked lists without changing links of first list for more details!

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Last Updated : 31 Mar, 2022
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