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 first list and rest elements should be part of second list. Sorted order should be maintained. We are not allowed to change pointers of first linked list.

For example,

Input: First List: 2->4->7->8->10 Second List: 1->3->12 Output: 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 first linked list. If we are allowed to change links, we can simply do something like merge of merge-sort algorithm. We assign first n smallest elements to the first linked list where n is the number of elements in first linked list and the rest to 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 first list.

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

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

Below is the implementation of this idea.

## C++

`// 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->next->data)` ` ` `{` ` ` `b = b->next;` ` ` `struct` `Node *ptr= b, *prev = NULL;` ` ` `// find mismatch by traversing the` ` ` `// second linked list once` ` ` `while` `(ptr && ptr->data < temp->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: "` `;` ` ` `printList(a);` ` ` `cout << ` `"Second List: "` `;` ` ` `printList(b);` ` ` `return` `0;` `}` |

## Python3

`# Python3 program to merge two sorted linked ` `# lists without using any extra space and ` `# without changing links of first list` `# Structure for a linked list node ` `class` `Node:` ` ` ` ` `def` `__init__(` `self` `):` ` ` ` ` `self` `.data ` `=` `0` ` ` `self` `.` `next` `=` `None` ` ` `# 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. ` `def` `push(head_ref, new_data):` ` ` `# Allocate node ` ` ` `new_node ` `=` `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` ` ` `return` `head_ref` `# Function to merge two sorted linked lists` `# LL1 and LL2 without using any extra space.` `def` `mergeLists(a, b):` ` ` `# Run till either one of a ` ` ` `# or b runs out` ` ` `while` `(a ` `and` `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` ` ` `a.data, b.data ` `=` `b.data, a.data` ` ` ` ` `temp ` `=` `b` ` ` ` ` `# To keep LL2 sorted, place first` ` ` `# element of LL2 at its correct place` ` ` `if` `(b.` `next` `and` `b.data > b.` `next` `.data):` ` ` `b ` `=` `b.` `next` ` ` `ptr ` `=` `b` ` ` `prev ` `=` `None` ` ` ` ` `# Find mismatch by traversing the` ` ` `# second linked list once` ` ` `while` `(ptr ` `and` `ptr.data < temp.data):` ` ` `prev ` `=` `ptr` ` ` `ptr ` `=` `ptr.` `next` ` ` ` ` `# Correct the pointers` ` ` `prev.` `next` `=` `temp` ` ` `temp.` `next` `=` `ptr` ` ` ` ` `# Move LL1 pointer to next element` ` ` `a ` `=` `a.` `next` ` ` `# Function to print the linked link` `def` `printList(head):` ` ` `while` `(head):` ` ` `print` `(head.data, end ` `=` `'->'` `)` ` ` `head ` `=` `head.` `next` ` ` ` ` `print` `(` `'NULL'` `)` ` ` `# Driver code` `if` `__name__` `=` `=` `'__main__'` `:` ` ` ` ` `a ` `=` `None` ` ` `a ` `=` `push(a, ` `10` `)` ` ` `a ` `=` `push(a, ` `8` `)` ` ` `a ` `=` `push(a, ` `7` `)` ` ` `a ` `=` `push(a, ` `4` `)` ` ` `a ` `=` `push(a, ` `2` `)` ` ` ` ` `b ` `=` `None` ` ` `b ` `=` `push(b, ` `12` `)` ` ` `b ` `=` `push(b, ` `3` `)` ` ` `b ` `=` `push(b, ` `1` `)` ` ` ` ` `mergeLists(a, b)` ` ` ` ` `print` `(` `"First List: "` `, end ` `=` `'')` ` ` `printList(a)` ` ` ` ` `print` `(` `"Second List: "` `, end ` `=` `'')` ` ` `printList(b)` `# This code is contributed by rutvik_56` |

**Output : **

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

**Time Complexity : **O(mn)

This article is contributed by **Aditya Goel**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.