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

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 <bits/stdc++.h>
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<Node*> 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 of 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

Time complexity: O(n*k) where n is size of given linked list

Auxiliary Space: O(1)

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