Reverse a singly Linked List in groups of given size | Set 3

Given a singly linked list and an integer K, the task is to reverse every K nodes of the given linked list.

Examples:

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



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

Approach: Two different approaches to solve this problem have been discussed in Set 1 and Set 2 of this article. In this article, an approach based on deque will be discussed.

  1. Create a deque.
  2. Store the address of the first k nodes in the deque.
  3. Pop first and the last value from the deque and swap the data values at those addresses.
  4. Repeat step 3 till the deque is not empty.
  5. Repeat step 2 for the next k nodes and till the end of the linked list is not reached.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Link list node
struct node {
    int data;
    struct node* next;
};
  
// Function to insert a node at
// the head of the linked list
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 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 the linked list
void printList(node* head)
{
    while (head != NULL) {
        cout << head->data << " ";
        head = head->next;
    }
}
  
/* Function to reverse the linked list in groups of 
size k and return the pointer to the new head node. */
struct node* reverse(struct node* head, int k)
{
  
    if (head == NULL)
        return head;
  
    // Create deque to store the address
    // of the nodes of the linked list
    deque<node*> q;
  
    // Store head pointer in current to
    // traverse the linked list
    node* current = head;
    int i;
  
    // Iterate through the entire linked
    // list by moving the current
    while (current != NULL) {
        i = 1;
  
        // Store addresses of the k
        // nodes in the deque
        while (i <= k) {
            if (current == NULL)
                break;
            q.push_back(current);
            current = current->next;
            i++;
        }
  
        /* pop first and the last value from 
        the deque and swap the data values at 
        those addresses
        Do this till there exist an address in 
        the deque or deque is not empty*/
        while (!q.empty()) {
            node* front = q.front();
            node* last = q.back();
            swap(front->data, last->data);
  
            // pop from the front if
            // the deque is not empty
            if (!q.empty())
                q.pop_front();
  
            // pop from the back if
            // the deque is not empty
            if (!q.empty())
                q.pop_back();
        }
    }
    return head;
}
  
// Driver code
int main()
{
  
    // Start with the empty list
    node* head = NULL;
  
    // Created Linked list is
    // 1->2->3->4->5->6->7->8->9->10
    push(&head, 10);
    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);
  
    int k = 2;
  
    // Get the new head after reversing the
    // linked list in groups of size k
    head = reverse(head, k);
    printList(head);
  
    return 0;
}

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Pyhton3

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# Python3 implementation of the approach
  
# Link list node
class Node:
    def __init__(self):
        self.data = 0
        self.next = None
  
# Function to insert a node at
# the head of the linked list
def push(head_ref, new_data):
  
    # Allocate node 
    new_node = 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
    return head_ref
  
# Function to print the linked list
def printList( head):
  
    while (head != None) :
        print( head.data, end = " ")
        head = head.next
      
# Function to reverse the linked list in groups of 
# size k and return the pointer to the new head node. 
def reverse( head, k):
  
    if (head == None):
        return head
  
    # Create deque to store the address
    # of the nodes of the linked list
    q = []
  
    # Store head pointer in current to
    # traverse the linked list
    current = head
    i = 0
  
    # Iterate through the entire linked
    # list by moving the current
    while (current != None) :
        i = 1
  
        # Store addresses of the k
        # nodes in the deque
        while (i <= k) :
            if (current == None):
                break
            q.append(current)
            current = current.next
            i = i + 1
          
        # pop first and the last value from 
        # the deque and swap the data values at 
        # those addresses
        # Do this till there exist an address in 
        # the deque or deque is not empty
        while (len(q) > 0): 
            front = q[-1]
            last = q[0]
              
            temp = front.data
            front.data = last.data
            last.data = temp
  
            # pop from the front if
            # the deque is not empty
            if (len(q) > 0):
                q.pop()
  
            # pop from the back if
            # the deque is not empty
            if (len(q)):
                q.pop(0)
      
    return head
  
# Driver code
  
# Start with the empty list
head = None
  
# Created Linked list is
# 1.2.3.4.5.6.7.8.9.10
head = push(head, 10)
head = push(head, 9)
head = push(head, 8)
head = push(head, 7)
head = push(head, 6)
head = push(head, 5)
head = push(head, 4)
head = push(head, 3)
head = push(head, 2)
head = push(head, 1)
  
k = 2
  
# Get the new head after reversing the
# linked list in groups of size k
head = reverse(head, k)
printList(head)
  
# This code is contributed by Arnab Kundu

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Output:

Given linked list 
1 2 3 4 5 6 7 8 9 10           
K=2    
Reversed Linked list
2 1 4 3 6 5 8 7 10 9

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