Merge two sorted linked lists such that merged list is in reverse order

Given two linked lists sorted in increasing order. Merge them such a way that the result list is in decreasing order (reverse order).

Examples: 

Input:  a: 5->10->15->40
        b: 2->3->20 
Output: res: 40->20->15->10->5->3->2

Input:  a: NULL
        b: 2->3->20 
Output: res: 20->3->2

A Simple Solution is to do following. 
1) Reverse first list ‘a’. 
2) Reverse second list ‘b’. 
3) Merge two reversed lists.
Another Simple Solution is first Merge both lists, then reverse the merged list.
Both of the above solutions require two traversals of linked list. 

How to solve without reverse, O(1) auxiliary space (in-place) and only one traversal of both lists?
The idea is to follow merge style process. Initialize result list as empty. Traverse both lists from beginning to end. Compare current nodes of both lists and insert smaller of two at the beginning of the result list. 

1) Initialize result list as empty: res = NULL.
2) Let 'a' and 'b' be heads first and second lists respectively.
3) While (a != NULL and b != NULL)
    a) Find the smaller of two (Current 'a' and 'b')
    b) Insert the smaller value node at the front of result.
    c) Move ahead in the list of smaller node. 
4) If 'b' becomes NULL before 'a', insert all nodes of 'a' 
   into result list at the beginning.
5) If 'a' becomes NULL before 'b', insert all nodes of 'a' 
   into result list at the beginning. 

Below is the implementation of above solution.

C++14

/* Given two sorted non-empty linked lists. Merge them in
   such a way that the result list will be in reverse
   order. Reversing of linked list is not allowed. Also,
   extra space should be O(1) */
#include<iostream>
using namespace std;

/* Link list Node */
struct Node
{
    int key;
    struct Node* next;
};

// Given two non-empty linked lists 'a' and 'b'
Node* SortedMerge(Node *a, Node *b)
{
    // If both lists are empty
    if (a==NULL && b==NULL) return NULL;

    // Initialize head of resultant list
    Node *res = NULL;

    // Traverse both lists while both of then
    // have nodes.
    while (a != NULL && b != NULL)
    {
        // If a's current value is smaller or equal to
        // b's current value.
        if (a->key <= b->key)
        {
            // Store next of current Node in first list
            Node *temp = a->next;

            // Add 'a' at the front of resultant list
            a->next = res;
            res = a;

            // Move ahead in first list
            a = temp;
        }

        // If a's value is greater. Below steps are similar
        // to above (Only 'a' is replaced with 'b')
        else
        {
            Node *temp = b->next;
            b->next = res;
            res = b;
            b = temp;
        }
    }

    // If second list reached end, but first list has
    // nodes. Add remaining nodes of first list at the
    // front of result list
    while (a != NULL)
    {
        Node *temp = a->next;
        a->next = res;
        res = a;
        a = temp;
    }

    // If first list reached end, but second list has
    // node. Add remaining nodes of first list at the
    // front of result list
    while (b != NULL)
    {
        Node *temp = b->next;
        b->next = res;
        res = b;
        b = temp;
    }

    return res;
}

/* Function to print Nodes in a given linked list */
void printList(struct Node *Node)
{
    while (Node!=NULL)
    {
        cout << Node->key << " ";
        Node = Node->next;
    }
}

/* Utility function to create a new node with
   given key */
Node *newNode(int key)
{
    Node *temp = new Node;
    temp->key = key;
    temp->next = NULL;
    return temp;
}

/* Driver program to test above functions*/
int main()
{
    /* Start with the empty list */
    struct Node* res = NULL;

    /* Let us create two sorted linked lists to test
       the above functions. Created lists shall be
         a: 5->10->15
         b: 2->3->20  */
    Node *a = newNode(5);
    a->next = newNode(10);
    a->next->next = newNode(15);

    Node *b = newNode(2);
    b->next = newNode(3);
    b->next->next = newNode(20);

    cout << "List A before merge: \n";
    printList(a);

    cout << "\nList B before merge: \n";
    printList(b);

    /* merge 2 increasing order LLs in decreasing order */
    res = SortedMerge(a, b);

    cout << "\nMerged Linked List is: \n";
    printList(res);

    return 0;
}

Java

// Java program to merge two sorted linked list such that merged
// list is in reverse order

// Linked List Class
class LinkedList {

    Node head;  // head of list
    static Node a, b;

    /* Node Class */
    static class Node {

        int data;
        Node next;

        // Constructor to create a new node
        Node(int d) {
            data = d;
            next = null;
        }
    }

    void printlist(Node node) {
        while (node != null) {
            System.out.print(node.data + " ");
            node = node.next;
        }
    }

    Node sortedmerge(Node node1, Node node2) {

        // if both the nodes are null
        if (node1 == null && node2 == null) {
            return null;
        }

        // resultant node
        Node res = null;

        // if both of them have nodes present traverse them
        while (node1 != null && node2 != null) {

            // Now compare both nodes current data
            if (node1.data <= node2.data) {
                Node temp = node1.next;
                node1.next = res;
                res = node1;
                node1 = temp;
            } else {
                Node temp = node2.next;
                node2.next = res;
                res = node2;
                node2 = temp;
            }
        }

        // If second list reached end, but first list has
        // nodes. Add remaining nodes of first list at the
        // front of result list
        while (node1 != null) {
            Node temp = node1.next;
            node1.next = res;
            res = node1;
            node1 = temp;
        }

        // If first list reached end, but second list has
        // node. Add remaining nodes of first list at the
        // front of result list
        while (node2 != null) {
            Node temp = node2.next;
            node2.next = res;
            res = node2;
            node2 = temp;
        }

        return res;

    }

    public static void main(String[] args) {

        LinkedList list = new LinkedList();
        Node result = null;

        /*Let us create two sorted linked lists to test
         the above functions. Created lists shall be
         a: 5->10->15
         b: 2->3->20*/
        list.a = new Node(5);
        list.a.next = new Node(10);
        list.a.next.next = new Node(15);

        list.b = new Node(2);
        list.b.next = new Node(3);
        list.b.next.next = new Node(20);

        System.out.println("List a before merge :");
        list.printlist(a);
        System.out.println("");
        System.out.println("List b before merge :");
        list.printlist(b);

        // merge two sorted linkedlist in decreasing order
        result = list.sortedmerge(a, b);
        System.out.println("");
        System.out.println("Merged linked list : ");
        list.printlist(result);

    }
}

// This code has been contributed by Mayank Jaiswal

Python3

# Given two sorted non-empty linked lists. Merge them in
# such a way that the result list will be in reverse
# order. Reversing of linked list is not allowed. Also,
# extra space should be O(1)

# Node of a linked list
class Node:
    def __init__(self, next = None, data = None):
        self.next = next
        self.data = data

# Given two non-empty linked lists 'a' and 'b'
def SortedMerge(a,b):

    # If both lists are empty
    if (a == None and b == None):
        return None

    # Initialize head of resultant list
    res = None

    # Traverse both lists while both of then
    # have nodes.
    while (a != None and b != None):

        # If a's current value is smaller or equal to
        # b's current value.
        if (a.key <= b.key):

            # Store next of current Node in first list
            temp = a.next

            # Add 'a' at the front of resultant list
            a.next = res
            res = a

            # Move ahead in first list
            a = temp

        # If a's value is greater. Below steps are similar
        # to above (Only 'a' is replaced with 'b')
        else:

            temp = b.next
            b.next = res
            res = b
            b = temp

    # If second list reached end, but first list has
    # nodes. Add remaining nodes of first list at the
    # front of result list
    while (a != None):

        temp = a.next
        a.next = res
        res = a
        a = temp

    # If first list reached end, but second list has
    # node. Add remaining nodes of first list at the
    # front of result list
    while (b != None):

        temp = b.next
        b.next = res
        res = b
        b = temp

    return res

# Function to print Nodes in a given linked list
def printList(Node):

    while (Node != None):

        print( Node.key, end = " ")
        Node = Node.next

# Utility function to create a new node with
# given key
def newNode( key):

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

# Driver program to test above functions

# Start with the empty list
res = None

# Let us create two sorted linked lists to test
# the above functions. Created lists shall be
#     a: 5.10.15
#     b: 2.3.20
a = newNode(5)
a.next = newNode(10)
a.next.next = newNode(15)

b = newNode(2)
b.next = newNode(3)
b.next.next = newNode(20)

print( "List A before merge: ")
printList(a)

print( "\nList B before merge: ")
printList(b)

# merge 2 increasing order LLs in decreasing order
res = SortedMerge(a, b)

print("\nMerged Linked List is: ")
printList(res)

# This code is contributed by Arnab Kundu

C#

// C# program to merge two sorted
// linked list such that merged
// list is in reverse order

// Linked List Class
using System;

class LinkedList
{

    public Node head; // head of list
    static Node a, b;

    /* Node Class */
    public class Node
    {
        public int data;
        public Node next;

        // Constructor to create a new node
        public Node(int d)
        {
            data = d;
            next = null;
        }
    }

    void printlist(Node node)
    {
        while (node != null)
        {
            Console.Write(node.data + " ");
            node = node.next;
        }
    }

    Node sortedmerge(Node node1, Node node2)
    {

        // if both the nodes are null
        if (node1 == null && node2 == null)
        {
            return null;
        }

        // resultant node
        Node res = null;

        // if both of them have nodes
        // present traverse them
        while (node1 != null && node2 != null)
        {

            // Now compare both nodes current data
            if (node1.data <= node2.data)
            {
                Node temp = node1.next;
                node1.next = res;
                res = node1;
                node1 = temp;
            }
            else
            {
                Node temp = node2.next;
                node2.next = res;
                res = node2;
                node2 = temp;
            }
        }

        // If second list reached end, but first
        // list has nodes. Add remaining nodes of
        // first list at the front of result list
        while (node1 != null)
        {
            Node temp = node1.next;
            node1.next = res;
            res = node1;
            node1 = temp;
        }

        // If first list reached end, but second
        // list has node. Add remaining nodes of
        // first list at the front of result list
        while (node2 != null)
        {
            Node temp = node2.next;
            node2.next = res;
            res = node2;
            node2 = temp;
        }

        return res;

    }

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

        LinkedList list = new LinkedList();
        Node result = null;

        /*Let us create two sorted linked lists to test
        the above functions. Created lists shall be
        a: 5->10->15
        b: 2->3->20*/
        LinkedList.a = new Node(5);
        LinkedList.a.next = new Node(10);
        LinkedList.a.next.next = new Node(15);

        LinkedList.b = new Node(2);
        LinkedList.b.next = new Node(3);
        LinkedList.b.next.next = new Node(20);

        Console.WriteLine("List a before merge :");
        list.printlist(a);
        Console.WriteLine("");
        Console.WriteLine("List b before merge :");
        list.printlist(b);

        // merge two sorted linkedlist in decreasing order
        result = list.sortedmerge(a, b);
        Console.WriteLine("");
        Console.WriteLine("Merged linked list : ");
        list.printlist(result);

    }
}

// This code has been contributed by 29AjayKumar

Javascript

<script>

// Javascript program to merge two
// sorted linked list such that merged
// list is in reverse order

// Node Class
class Node
{
    constructor(d)
    {
        this.data = d;
        this.next = null;
    }
}

// Head of list
let head;
let a, b;

function printlist(node)
{
    while (node != null)
    {
        document.write(node.data + " ");
        node = node.next;
    }
}

function sortedmerge(node1, node2)
{

    // If both the nodes are null
    if (node1 == null && node2 == null)
    {
        return null;
    }

    // Resultant node
    let res = null;

    // If both of them have nodes present
    // traverse them
    while (node1 != null && node2 != null)
    {

        // Now compare both nodes current data
        if (node1.data <= node2.data)
        {
            let temp = node1.next;
            node1.next = res;
            res = node1;
            node1 = temp;
        }
        else
        {
            let temp = node2.next;
            node2.next = res;
            res = node2;
            node2 = temp;
        }
    }

    // If second list reached end, but
    // first list has nodes. Add
    // remaining nodes of first
    // list at the front of result list
    while (node1 != null)
    {
        let temp = node1.next;
        node1.next = res;
        res = node1;
        node1 = temp;
    }

    // If first list reached end, but
    // second list has node. Add
    // remaining nodes of first
    // list at the front of result list
    while (node2 != null)
    {
        let temp = node2.next;
        node2.next = res;
        res = node2;
        node2 = temp;
    }
    return res;
}

// Driver code
let result = null;

/*Let us create two sorted linked lists to test
         the above functions. Created lists shall be
         a: 5->10->15
         b: 2->3->20*/
a = new Node(5);
a.next = new Node(10);
a.next.next = new Node(15);

b = new Node(2);
b.next = new Node(3);
b.next.next = new Node(20);

document.write("List a before merge :<br>");
printlist(a);
document.write("<br>");
document.write("List b before merge :<br>");
printlist(b);

// Merge two sorted linkedlist in decreasing order
result = sortedmerge(a, b);
document.write("<br>");
document.write("Merged linked list : <br>");
printlist(result);

// This code is contributed by rag2127

</script>

Output

List A before merge: 
5 10 15 
List B before merge: 
2 3 20 
Merged Linked List is: 
20 15 10 5 3 2 

Time Complexity: O(N)
Auxiliary Space: O(1)

Another Approach:

1. Define a struct for linked list nodes with two fields: “data” and “next”.
2. Define a function “mergeListsReverse” that takes two linked lists “a” and “b” as inputs and returns a pointer to the head of the merged linked list.
3. Initialize two pointers “head” and “tail” to “nullptr” for the merged linked list.
4. Use a while loop to traverse both linked lists “a” and “b” simultaneously.
5. Compare the data values of the current nodes in “a” and “b”, and append the smaller node to the head of the merged list.
6. Update the corresponding pointer to the next node in the linked list that was appended to the merged list.
7. Continue this process until one of the input lists is fully traversed.
8. If there are any remaining nodes in list “a”, append them to the merged list.
9. If there are any remaining nodes in list “b”, append them to the merged list.
10. Return the head pointer of the merged list.
11. Define two utility functions:
    a. “push”: takes a pointer to a pointer to the head of a linked list and a new integer value, and inserts a new node with the value at the beginning of the list.
    b. “printList”: takes a pointer to the head of a linked list and prints the values of all nodes in the list.
    In the main function, create two empty linked lists “headA” and “headB”.
12. Use the “push” function to insert several integer values into each of the linked lists.
13. Print the values of the two input linked lists using the “printList” function.
14. Call the “mergeListsReverse” function with the two input linked lists as arguments, and assign the returned head pointer to a new pointer variable “merged”.
15. Print the values of the merged linked list using the “printList” function.
    

C++

#include <iostream>

using namespace std;

/* Linked list node */
struct Node {
    int data;
    Node* next;
};

/* Given two non-empty linked lists 'a' and 'b',
   merge them in such a way that the result list
   will be in reverse order */
Node* mergeListsReverse(Node* a, Node* b) {
    Node* head = nullptr;  // head of the merged list
    Node* tail = nullptr;  // tail of the merged list

    while (a && b) {
        if (a->data < b->data) {
            Node* temp = a->next;
            a->next = head;
            head = a;
            a = temp;
        } else {
            Node* temp = b->next;
            b->next = head;
            head = b;
            b = temp;
        }
    }

    while (a) {
        Node* temp = a->next;
        a->next = head;
        head = a;
        a = temp;
    }

    while (b) {
        Node* temp = b->next;
        b->next = head;
        head = b;
        b = temp;
    }

    return head;
}

/* Utility function to insert a node at the beginning of a linked list */
void push(Node** headRef, int newData) {
    Node* newNode = new Node();
    newNode->data = newData;
    newNode->next = *headRef;
    *headRef = newNode;
}

/* Utility function to print a linked list */
void printList(Node* head) {
    while (head) {
        cout << head->data << " ";
        head = head->next;
    }
    cout << endl;
}

/* Driver program to test above functions */
int main() {
    Node* headA = nullptr;
    Node* headB = nullptr;

    // Populate list A
    push(&headA, 15);
    push(&headA, 10);
    push(&headA, 5);

    // Populate list B
    push(&headB, 20);
    push(&headB, 3);
    push(&headB, 2);

    cout << "List A before merge: ";
    printList(headA);

    cout << "List B before merge: ";
    printList(headB);

    // Merge the two lists in reverse order
    Node* merged = mergeListsReverse(headA, headB);

    cout << "Merged list in reverse order: ";
    printList(merged);

    return 0;
}

C#

using System;

// Linked list node
public class Node {
    public int data;
    public Node next;
}

public class Program {
    // Given two non-empty linked lists 'a' and 'b',
    // merge them in such a way that the result list
    // will be in reverse order
    public static Node MergeListsReverse(Node a, Node b) {
        Node head = null;  // head of the merged list
        Node tail = null;  // tail of the merged list

        while (a != null && b != null) {
            if (a.data < b.data) {
                Node temp = a.next;
                a.next = head;
                head = a;
                a = temp;
            } else {
                Node temp = b.next;
                b.next = head;
                head = b;
                b = temp;
            }
        }

        while (a != null) {
            Node temp = a.next;
            a.next = head;
            head = a;
            a = temp;
        }

        while (b != null) {
            Node temp = b.next;
            b.next = head;
            head = b;
            b = temp;
        }

        return head;
    }

    // Utility function to insert a node at the beginning of a linked list
    public static void Push(ref Node headRef, int newData) {
        Node newNode = new Node();
        newNode.data = newData;
        newNode.next = headRef;
        headRef = newNode;
    }

    // Utility function to print a linked list
    public static void PrintList(Node head) {
        while (head != null) {
            Console.Write(head.data + " ");
            head = head.next;
        }
        Console.WriteLine();
    }

    // Driver program to test above functions
    public static void Main() {
        Node headA = null;
        Node headB = null;

        // Populate list A
        Push(ref headA, 15);
        Push(ref headA, 10);
        Push(ref headA, 5);

        // Populate list B
        Push(ref headB, 20);
        Push(ref headB, 3);
        Push(ref headB, 2);

        Console.Write("List A before merge: ");
        PrintList(headA);

        Console.Write("List B before merge: ");
        PrintList(headB);

        // Merge the two lists in reverse order
        Node merged = MergeListsReverse(headA, headB);

        Console.Write("Merged list in reverse order: ");
        PrintList(merged);
    }
}

Java

class Node {
    int data;
    Node next;

    public Node(int data) {
        this.data = data;
        this.next = null;
    }
}

public class Main {
    /* Given two non-empty linked lists 'a' and 'b',
    merge them in such a way that the result list
    will be in reverse order */
    public static Node mergeListsReverse(Node a, Node b) {
        Node head = null; // head of the merged list
        Node tail = null; // tail of the merged list

        while (a != null && b != null) {
            if (a.data < b.data) {
                Node temp = a.next;
                a.next = head;
                head = a;
                a = temp;
            } else {
                Node temp = b.next;
                b.next = head;
                head = b;
                b = temp;
            }
        }

        while (a != null) {
            Node temp = a.next;
            a.next = head;
            head = a;
            a = temp;
        }

        while (b != null) {
            Node temp = b.next;
            b.next = head;
            head = b;
            b = temp;
        }

        return head;
    }

    /* Utility function to insert a node at the beginning of a linked list */
    public static void push(Node headRef, int newData) {
        Node newNode = new Node(newData);
        newNode.next = headRef;
        headRef = newNode;
    }

    /* Utility function to print a linked list */
    public static void printList(Node head) {
        while (head != null) {
            System.out.print(head.data + " ");
            head = head.next;
        }
        System.out.println();
    }

    /* Driver program to test above functions */
    public static void main(String[] args) {
        Node headA = null;
        Node headB = null;

        // Populate list A
        headA = new Node(5);
        headA.next = new Node(10);
        headA.next.next = new Node(15);

        // Populate list B
        headB = new Node(2);
        headB.next = new Node(3);
        headB.next.next = new Node(20);

        System.out.print("List A before merge: ");
        printList(headA);

        System.out.print("List B before merge: ");
        printList(headB);

        // Merge the two lists in reverse order
        Node merged = mergeListsReverse(headA, headB);

        System.out.print("Merged list in reverse order: ");
        printList(merged);
    }
}
// This code is contributed by rudra1807raj

Output

List A before merge: 5 10 15 
List B before merge: 2 3 20 
Merged list in reverse order: 20 15 10 5 3 2 

Time Complexity: The time complexity of the mergeListsReverse() function is O(m+n), where m and n are the lengths of the input linked lists ‘a’ and ‘b’, respectively. This is because each node in both lists is visited exactly once, and constant time operations are performed on each node.

Auxiliary Space: The space complexity of the mergeListsReverse() function is O(1), which is constant space, as only a few temporary variables are used to store the links between the nodes. The space required does not depend on the length of the input lists. However, if we consider the space required to store the linked lists themselves, then the space complexity is O(m+n), where m and n are the lengths of the input linked lists. This is because we are creating a new linked list that has all the elements of both input lists.

This solution traverses both lists only once, doesn’t require reverse and works in-place.
This article is contributed by Mohammed Raqeeb. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.