# Reverse a Linked List in groups of given size | Set 1

• Difficulty Level : Medium
• Last Updated : 08 Jan, 2022

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

• 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:

## Python3

 # Python program to reverse a# linked list in group of given size # Node class  class Node:     # Constructor to initialize the node object    def __init__(self, data):        self.data = data        self.next = None  class LinkedList:     # Function to initialize head    def __init__(self):        self.head = None     def reverse(self, head, k):               if head == None:          return None        current = head        next = None        prev = None        count = 0         # Reverse first k nodes of the linked list        while(current is not None and count < k):            next = current.next            current.next = prev            prev = current            current = next            count += 1         # 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 is not None:            head.next = self.reverse(next, k)         # prev is new head of the input list        return prev     # Function to insert a new node at the beginning    def push(self, new_data):        new_node = Node(new_data)        new_node.next = self.head        self.head = new_node     # Utility function to print the linked LinkedList    def printList(self):        temp = self.head        while(temp):            print(temp.data,end=' ')            temp = temp.next  # Driver programllist = LinkedList()llist.push(9)llist.push(8)llist.push(7)llist.push(6)llist.push(5)llist.push(4)llist.push(3)llist.push(2)llist.push(1) print("Given linked list")llist.printList()llist.head = llist.reverse(llist.head, 3) print ("\nReversed Linked list")llist.printList() # This code is contributed by Nikhil Kumar Singh(nickzuck_007)

## Javascript



Output:

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 write comments if you find the above code/algorithm incorrect, or find other ways to solve the same problem.

My Personal Notes arrow_drop_up