C++ Program For Reversing A Linked List In Groups Of Given Size – Set 1

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

Example:

Input: 1->2->3->4->5->6->7->8->NULL, K = 3 
Output: 3->2->1->6->5->4->8->7->NULL 
Input: 1->2->3->4->5->6->7->8->NULL, K = 5 
Output: 5->4->3->2->1->8->7->6->NULL 

Algorithm: reverse(head, k)

• Reverse the first sub-list of size k. While reversing keep track of the next node and previous node. Let the pointer to the next node be next and pointer to the previous node be prev. See this post for reversing a linked list.
• head->next = reverse(next, k) ( Recursively call for rest of the list and link the two sub-lists )
• Return prev ( prev becomes the new head of the list (see the diagrams of an iterative method of this post )

Below is image shows how the reverse function works: 

Below is the implementation of the above approach:

C++

// C++ program to reverse a linked list
// in groups of given size
#include <bits/stdc++.h>
using namespace std;

// Link list node
class Node
{
    public:
    int data;
    Node* next;
};

/* Reverses the linked list in groups
  of size k and returns the pointer
  to the new head node. */
Node* reverse(Node* head, int k)
{
    // Base case
    if (!head)
        return NULL;
    Node* current = head;
    Node* next = NULL;
    Node* prev = NULL;
    int count = 0;

    // Reverse first k nodes of the
    // linked list
    while (current != NULL &&
           count < k)
    {
        next = current->next;
        current->next = prev;
        prev = current;
        current = next;
        count++;
    }

    /* next is now a pointer to (k+1)th node
       Recursively call for the list starting
       from current. And make rest of the list
       as next of first node */
    if (next != NULL)
        head->next = reverse(next, k);

    // prev is new head of the input list
    return prev;
}

// UTILITY FUNCTIONS
// Function to push a node
void push(Node** head_ref,
          int new_data)
{
    // Allocate node
    Node* new_node = new 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(Node* node)
{
    while (node != NULL)
    {
        cout << node->data << " ";
        node = node->next;
    }
}

// Driver code
int main()
{
    // Start with the empty list
    Node* head = NULL;

    /* Create Linked list
       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);

    cout << "Given linked list ";
    printList(head);
    head = reverse(head, 3);

    cout << "Reversed Linked list ";
    printList(head);

    return (0);
}
// This code is contributed by rathbhupendra

Output:

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

Complexity Analysis:

• Time Complexity: O(n). 
  Traversal of list is done only once and it has ‘n’ elements.
• Auxiliary Space: O(n/k). 
  For each Linked List of size n, n/k or (n/k)+1 calls will be made during the recursion.

Please refer complete article on Reverse a Linked List in groups of given size | Set 1 for more details!