Reverse a Linked List in groups of given size | Set 2

3

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

Examples:

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

Inputs:   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++ 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 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;
    }
}

/* Drier program to test above function*/
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("\nGiven linked list \n");
    printList(head);
    head = Reverse(head, 3);

    printf("\nReversed Linked list \n");
    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

This article is contributed by Jatin Goyal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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