# C++ Program For Reversing A Linked List In Groups Of Given Size – Set 2

• Difficulty Level : Hard
• Last Updated : 11 Jan, 2022

Given a linked list, write a function to reverse every k nodes (where k is an input to the function).
Examples:

```Input: 1->2->3->4->5->6->7->8->NULL and k = 3
Output: 3->2->1->6->5->4->8->7->NULL.

Input: 1->2->3->4->5->6->7->8->NULL and k = 5
Output: 5->4->3->2->1->8->7->6->NULL.```

We have already discussed its solution in below post
Reverse a Linked List in groups of given size | Set 1
In this post, we have used a stack which will store the nodes of the given linked list. Firstly, push the k elements of the linked list in the stack. Now pop elements one by one and keep track of the previously popped node. Point the next pointer of prev node to top element of stack. Repeat this process, until NULL is reached.
This algorithm uses O(k) extra space.

## C++

 `// C++ program to reverse a linked list ``// in groups of given size``#include ``using` `namespace` `std;`` ` `// Link list node``struct` `Node ``{``    ``int` `data;``    ``struct` `Node* next;``};`` ` `/* Reverses the linked list in groups ``   ``of size k and returns the pointer ``   ``to the new head node. */``struct` `Node* Reverse(``struct` `Node* head, ``                     ``int` `k)``{``    ``// Create a stack of Node*``    ``stack mystack;``    ``struct` `Node* current = head;``    ``struct` `Node* prev = NULL;`` ` `    ``while` `(current != NULL) ``    ``{``        ``// Terminate the loop whichever ``        ``// comes first either current == NULL ``        ``// or count >= k``        ``int` `count = 0;``        ``while` `(current != NULL && ``               ``count < k) ``        ``{``            ``mystack.push(current);``            ``current = current->next;``            ``count++;``        ``}`` ` `        ``// Now pop the elements of stack ``        ``// one by one``        ``while` `(mystack.size() > 0) ``        ``{``            ``// If final list has not been ``            ``// started yet.``            ``if` `(prev == NULL) ``            ``{``                ``prev = mystack.top();``                ``head = prev;``                ``mystack.pop();``            ``} ``            ``else` `            ``{``                ``prev->next = mystack.top();``                ``prev = prev->next;``                ``mystack.pop();``            ``}``        ``}``    ``}`` ` `    ``// Next of last element will ``    ``// point to NULL.``    ``prev->next = NULL;`` ` `    ``return` `head;``}`` ` `// UTILITY FUNCTIONS``// Function to push a node``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 print linked list``void` `printList(``struct` `Node* node)``{``    ``while` `(node != NULL) ``    ``{``        ``printf``(``"%d  "``, node->data);``        ``node = node->next;``    ``}``}`` ` `// Driver code``int` `main(``void``)``{``    ``// Start with the empty list``    ``struct` `Node* head = NULL;`` ` `    ``// Created Linked list is ``    ``// 1->2->3->4->5->6->7->8->9 ``    ``push(&head, 9);``    ``push(&head, 8);``    ``push(&head, 7);``    ``push(&head, 6);``    ``push(&head, 5);``    ``push(&head, 4);``    ``push(&head, 3);``    ``push(&head, 2);``    ``push(&head, 1);`` ` `    ``printf``(``"Given linked list "``);``    ``printList(head);``    ``head = Reverse(head, 3);`` ` `    ``printf``(``"Reversed Linked list "``);``    ``printList(head);`` ` `    ``return` `0;``}`

Output:

```Given Linked List
1 2 3 4 5 6 7 8 9
Reversed list
3 2 1 6 5 4 9 8 7```

Please refer complete article on Reverse a Linked List in groups of given size | Set 2 for more details!

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