Merge two sorted lists (in-place)

• Difficulty Level : Medium
• Last Updated : 12 Nov, 2021

Given two sorted lists, merge them so as to produce a combined sorted list (without using extra space).
Examples:

Output : 4->5->6->7->8->9
Explanation: The output list is in sorted order.

Output : 1->2->3->4->5->7
Explanation: The output list is in sorted order.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

There are different discussed different solutions in post below.

Method 1 (Recursive)

Approach: The recursive solution can be formed, given the linked lists are sorted.

2. Find the smaller node among the two head nodes. The current element will be the smaller node among two head nodes.
3. The rest elements of both lists will appear after that.
4. Now run a recursive function with parameters, the next node of the smaller element, and the other head.
5. The recursive function will return the next smaller element linked with rest of the sorted element. Now point the next of current element to that, i.e curr_ele->next=recursivefunction()
6. Handle some corner cases.
• If both the heads are NULL return null.
• If one head is null return the other.

C++

// C program to merge two sorted linked lists
// in-place.
#include <bits/stdc++.h>
using namespace std;

struct Node {
int data;
struct Node* next;
};

// Function to create newNode in a linkedlist
Node* newNode(int key)
{
struct Node* temp = new Node;
temp->data = key;
temp->next = NULL;
return temp;
}

// A utility function to print linked list
void printList(Node* node)
{
while (node != NULL) {
printf("%d  ", node->data);
node = node->next;
}
}

// Merges two given lists in-place. This function
// mainly compares head nodes and calls mergeUtil()
Node* merge(Node* h1, Node* h2)
{
if (!h1)
return h2;
if (!h2)
return h1;

// whose head data is the least
if (h1->data < h2->data) {
h1->next = merge(h1->next, h2);
return h1;
}
else {
h2->next = merge(h1, h2->next);
return h2;
}
}

// Driver program
int main()
{

return 0;
}

Java

// Java program to merge two sorted
class GFG {

static class Node {
int data;
Node next;
};

// Function to create newNode in a linkedlist
static Node newNode(int key)
{
Node temp = new Node();
temp.data = key;
temp.next = null;
return temp;
}

// A utility function to print linked list
static void printList(Node node)
{
while (node != null) {
System.out.printf("%d ", node.data);
node = node.next;
}
}

// Merges two given lists in-place. This function
// mainly compares head nodes and calls mergeUtil()
static Node merge(Node h1, Node h2)
{
if (h1 == null)
return h2;
if (h2 == null)
return h1;

// whose head data is the least
if (h1.data < h2.data) {
h1.next = merge(h1.next, h2);
return h1;
}
else {
h2.next = merge(h1, h2.next);
return h2;
}
}

// Driver program
public static void main(String args[])
{

}
}

// This code is contributed by Arnab Kundu

Python3

# Python3 program to merge two
import math

class Node:
def __init__(self, data):
self.data = data
self.next = None

# Function to create newNode in a linkedlist
def newNode( key):
temp = Node(key)
temp.data = key
temp.next = None
return temp

# A utility function to print linked list
def printList( node):
while (node != None):
print(node.data, end = " ")
node = node.next

# Merges two given lists in-place.
# This function mainly compares
# head nodes and calls mergeUtil()
def merge( h1, h2):
if (h1 == None):
return h2
if (h2 == None):
return h1

# whose head data is the least
if (h1.data < h2.data):
h1.next = merge(h1.next, h2)
return h1

else:
h2.next = merge(h1, h2.next)
return h2

# Driver Code
if __name__=='__main__':

# This code is contributed by Srathore

C#

// C# program to merge two sorted
using System;

class GFG {

public class Node {
public int data;
public Node next;
};

// Function to create newNode in a linkedlist
static Node newNode(int key)
{
Node temp = new Node();
temp.data = key;
temp.next = null;
return temp;
}

// A utility function to print linked list
static void printList(Node node)
{
while (node != null) {
Console.Write("{0} ", node.data);
node = node.next;
}
}

// Merges two given lists in-place. This function
// mainly compares head nodes and calls mergeUtil()
static Node merge(Node h1, Node h2)
{
if (h1 == null)
return h2;
if (h2 == null)
return h1;

// whose head data is the least
if (h1.data < h2.data) {
h1.next = merge(h1.next, h2);
return h1;
}
else {
h2.next = merge(h1, h2.next);
return h2;
}
}

// Driver code
public static void Main(String[] args)
{

}
}

// This code has been contributed by 29AjayKumar

Javascript

<script>
// javascript program to merge two sorted
class Node {
constructor(){
this.data = 0;
this.next = null;
}
}

// Function to create newNode in a linkedlist
function newNode(key) {
temp = new Node();
temp.data = key;
temp.next = null;
return temp;
}

// A utility function to print linked list
function printList( node) {
while (node != null) {
document.write(node.data+" ");
node = node.next;
}
}

// Merges two given lists in-place. This function
// mainly compares head nodes and calls mergeUtil()
function merge( h1,  h2) {
if (h1 == null)
return h2;
if (h2 == null)
return h1;

// whose head data is the least
if (h1.data < h2.data) {
h1.next = merge(h1.next, h2);
return h1;
} else {
h2.next = merge(h1, h2.next);
return h2;
}
}

// Driver program

// This code contributed by umadevi9616
</script>

Output:

0 1 2 3 4 5

Complexity Analysis:

• Time complexity:O(n).
Only one traversal of the linked lists are needed.
• Auxiliary Space:O(n).
If the recursive stack space is taken into consideration.

Method 2 (Iterative)

Approach: This approach is very similar to the above recursive approach.

1. Traverse the list from start to end.
2. If the head node of second list lies in between two nodes of the first list, insert it there and make the next node of second list the head. Continue this until there is no node left in both lists, i.e. both the lists are traversed.
3. If the first list has reached end while traversing, point the next node to the head of second list.

Note: Compare both the lists where the list with a smaller head value is the first list.

C++

// C++ program to merge two sorted linked lists
// in-place.
#include <bits/stdc++.h>
using namespace std;

struct Node {
int data;
struct Node* next;
};

// Function to create newNode in a linkedlist
struct Node* newNode(int key)
{
struct Node* temp = new Node;
temp->data = key;
temp->next = NULL;
return temp;
}

// A utility function to print linked list
void printList(struct Node* node)
{
while (node != NULL) {
printf("%d  ", node->data);
node = node->next;
}
}

// Merges two lists with headers as h1 and h2.
// It assumes that h1's data is smaller than
// or equal to h2's data.
struct Node* mergeUtil(struct Node* h1,
struct Node* h2)
{
// if only one node in first list
// simply point its head to second list
if (!h1->next) {
h1->next = h2;
return h1;
}

// Initialize current and next pointers of
// both lists
struct Node *curr1 = h1, *next1 = h1->next;
struct Node *curr2 = h2, *next2 = h2->next;

while (next1 && curr2) {
// if curr2 lies in between curr1 and next1
// then do curr1->curr2->next1
if ((curr2->data) >= (curr1->data) && (curr2->data) <= (next1->data)) {
next2 = curr2->next;
curr1->next = curr2;
curr2->next = next1;

// now let curr1 and curr2 to point
// to their immediate next pointers
curr1 = curr2;
curr2 = next2;
}
else {
// if more nodes in first list
if (next1->next) {
next1 = next1->next;
curr1 = curr1->next;
}

// else point the last node of first list
// to the remaining nodes of second list
else {
next1->next = curr2;
return h1;
}
}
}
return h1;
}

// Merges two given lists in-place. This function
// mainly compares head nodes and calls mergeUtil()
struct Node* merge(struct Node* h1,
struct Node* h2)
{
if (!h1)
return h2;
if (!h2)
return h1;

// whose head data is the least
if (h1->data < h2->data)
return mergeUtil(h1, h2);
else
return mergeUtil(h2, h1);
}

// Driver program
int main()
{

return 0;
}

Java

// Java program to merge two sorted
class GfG {

static class Node {
int data;
Node next;
}

// Function to create newNode in a linkedlist
static Node newNode(int key)
{
Node temp = new Node();
temp.data = key;
temp.next = null;
return temp;
}

// A utility function to print linked list
static void printList(Node node)
{
while (node != null) {
System.out.print(node.data + " ");
node = node.next;
}
}

// Merges two lists with headers as h1 and h2.
// It assumes that h1's data is smaller than
// or equal to h2's data.
static Node mergeUtil(Node h1, Node h2)
{
// if only one node in first list
// simply point its head to second list
if (h1.next == null) {
h1.next = h2;
return h1;
}

// Initialize current and next pointers of
// both lists
Node curr1 = h1, next1 = h1.next;
Node curr2 = h2, next2 = h2.next;

while (next1 != null && curr2 != null) {
// if curr2 lies in between curr1 and next1
// then do curr1->curr2->next1
if ((curr2.data) >= (curr1.data) && (curr2.data) <= (next1.data)) {
next2 = curr2.next;
curr1.next = curr2;
curr2.next = next1;

// now let curr1 and curr2 to point
// to their immediate next pointers
curr1 = curr2;
curr2 = next2;
}
else {
// if more nodes in first list
if (next1.next != null) {
next1 = next1.next;
curr1 = curr1.next;
}

// else point the last node of first list
// to the remaining nodes of second list
else {
next1.next = curr2;
return h1;
}
}
}
return h1;
}

// Merges two given lists in-place. This function
// mainly compares head nodes and calls mergeUtil()
static Node merge(Node h1, Node h2)
{
if (h1 == null)
return h2;
if (h2 == null)
return h1;

// whose head data is the least
if (h1.data < h2.data)
return mergeUtil(h1, h2);
else
return mergeUtil(h2, h1);
}

// Driver code
public static void main(String[] args)
{

}
}

// This code is contributed by
// prerna saini

Python

# Python program to merge two sorted linked lists
# in-place.

class Node:
def __init__(self, data):
self.data = data
self.next = None

# Function to create newNode in a linkedlist
def newNode(key):

temp = Node(0)
temp.data = key
temp.next = None
return temp

# A utility function to print linked list
def printList(node):

while (node != None) :
print( node.data, end =" ")
node = node.next

# Merges two lists with headers as h1 and h2.
# It assumes that h1's data is smaller than
# or equal to h2's data.
def mergeUtil(h1, h2):

# if only one node in first list
# simply point its head to second list
if (h1.next == None) :
h1.next = h2
return h1

# Initialize current and next pointers of
# both lists
curr1 = h1
next1 = h1.next
curr2 = h2
next2 = h2.next

while (next1 != None and curr2 != None):

# if curr2 lies in between curr1 and next1
# then do curr1.curr2.next1
if ((curr2.data) >= (curr1.data) and
(curr2.data) <= (next1.data)) :
next2 = curr2.next
curr1.next = curr2
curr2.next = next1

# now let curr1 and curr2 to point
# to their immediate next pointers
curr1 = curr2
curr2 = next2

else :
# if more nodes in first list
if (next1.next) :
next1 = next1.next
curr1 = curr1.next

# else point the last node of first list
# to the remaining nodes of second list
else :
next1.next = curr2
return h1

return h1

# Merges two given lists in-place. This function
# mainly compares head nodes and calls mergeUtil()
def merge( h1, h2):

if (h1 == None):
return h2
if (h2 == None):
return h1

# whose head data is the least
if (h1.data < h2.data):
return mergeUtil(h1, h2)
else:
return mergeUtil(h2, h1)

# Driver program

# This code is contributed by Arnab Kundu

C#

// C# program to merge two sorted
using System;

class GfG {

public class Node {
public int data;
public Node next;
}

// Function to create newNode in a linkedlist
static Node newNode(int key)
{
Node temp = new Node();
temp.data = key;
temp.next = null;
return temp;
}

// A utility function to print linked list
static void printList(Node node)
{
while (node != null) {
Console.Write(node.data + " ");
node = node.next;
}
}

// Merges two lists with headers as h1 and h2.
// It assumes that h1's data is smaller than
// or equal to h2's data.
static Node mergeUtil(Node h1, Node h2)
{
// if only one node in first list
// simply point its head to second list
if (h1.next == null) {
h1.next = h2;
return h1;
}

// Initialize current and next pointers of
// both lists
Node curr1 = h1, next1 = h1.next;
Node curr2 = h2, next2 = h2.next;

while (next1 != null && curr2 != null) {
// if curr2 lies in between curr1 and next1
// then do curr1->curr2->next1
if ((curr2.data) >= (curr1.data)
&& (curr2.data) <= (next1.data)) {
next2 = curr2.next;
curr1.next = curr2;
curr2.next = next1;

// now let curr1 and curr2 to point
// to their immediate next pointers
curr1 = curr2;
curr2 = next2;
}
else {
// if more nodes in first list
if (next1.next != null) {
next1 = next1.next;
curr1 = curr1.next;
}

// else point the last node of first list
// to the remaining nodes of second list
else {
next1.next = curr2;
return h1;
}
}
}
return h1;
}

// Merges two given lists in-place. This function
// mainly compares head nodes and calls mergeUtil()
static Node merge(Node h1, Node h2)
{
if (h1 == null)
return h2;
if (h2 == null)
return h1;

// whose head data is the least
if (h1.data < h2.data)
return mergeUtil(h1, h2);
else
return mergeUtil(h2, h1);
}

// Driver code
public static void Main(String[] args)
{

}
}

/* This code contributed by PrinciRaj1992 */

Javascript

<script>

// JavaScript program to merge two sorted
class Node {
constructor() {
this.data = 0;
this.next = null;
}
}

// Function to create newNode in a linkedlist
function newNode(key) {
var temp = new Node();
temp.data = key;
temp.next = null;
return temp;
}

// A utility function to print linked list
function printList(node) {
while (node != null) {
document.write(node.data + " ");
node = node.next;
}
}

// Merges two lists with headers as h1 and h2.
// It assumes that h1's data is smaller than
// or equal to h2's data.
function mergeUtil(h1,  h2) {
// if only one node in first list
// simply point its head to second list
if (h1.next == null) {
h1.next = h2;
return h1;
}

// Initialize current and next pointers of
// both lists
var curr1 = h1, next1 = h1.next;
var curr2 = h2, next2 = h2.next;

while (next1 != null && curr2 != null) {
// if curr2 lies in between curr1 and next1
// then do curr1->curr2->next1
if ((curr2.data) >= (curr1.data) &&
(curr2.data) <= (next1.data)) {
next2 = curr2.next;
curr1.next = curr2;
curr2.next = next1;

// now let curr1 and curr2 to point
// to their immediate next pointers
curr1 = curr2;
curr2 = next2;
} else {
// if more nodes in first list
if (next1.next != null) {
next1 = next1.next;
curr1 = curr1.next;
}

// else point the last node of first list
// to the remaining nodes of second list
else {
next1.next = curr2;
return h1;
}
}
}
return h1;
}

// Merges two given lists in-place. This function
// mainly compares head nodes and calls mergeUtil()
function merge(h1,  h2) {
if (h1 == null)
return h2;
if (h2 == null)
return h1;

// whose head data is the least
if (h1.data < h2.data)
return mergeUtil(h1, h2);
else
return mergeUtil(h2, h1);
}

// Driver code

// This code contributed by gauravrajput1

</script>

Output:

0 1 2 3 4 5

Complexity Analysis:

• Time complexity:O(n).
As only one traversal of the linked lists is needed.
• Auxiliary Space:O(1).
As there is no space required.

This article is contributed by Mandula Vikitha. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.