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# 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``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

 ``

Output

```List A before merge:
5 10 15
List B before merge:
2 3 20
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.
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 ` `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.