# Reverse 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

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

• 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 iterative method of this post) )

Below is image shows how the reverse function works: Below is the implementation of the above approach:

## C++

 `// CPP program to reverse a linked list ` `// in groups of given size  ` `#include ` `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)  ` `{  ` `    ``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 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(Node *node)  ` `{  ` `    ``while` `(node != NULL)  ` `    ``{  ` `        ``cout<data<<``" "``;  ` `        ``node = node->next;  ` `    ``}  ` `}  ` ` `  `/* 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 */` `    ``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 \n"``;  ` `    ``printList(head);  ` `    ``head = reverse(head, 3);  ` ` `  `    ``cout<<``"\nReversed Linked list \n"``;  ` `    ``printList(head);  ` ` `  `    ``return``(0);  ` `}  ` ` `  `// This code is contributed by rathbhupendra `

## C

 `// C program to reverse a linked list in groups of given size ` `#include ` `#include ` ` `  `/* 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) ` `{ ` `    ``struct` `Node* current = head; ` `    ``struct` `Node* next = NULL; ` `    ``struct` `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(``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; ` `    ``} ` `}     ` ` `  `/* Driver 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); ` `} `

## Java

 `// Java program to reverse a linked list in groups of ` `// given size ` `class` `LinkedList ` `{ ` `    ``Node head;  ``// head of list ` `  `  `    ``/* Linked list Node*/` `    ``class` `Node ` `    ``{ ` `        ``int` `data; ` `        ``Node next; ` `        ``Node(``int` `d) {data = d; next = ``null``; } ` `    ``} ` ` `  `    ``Node reverse(Node head, ``int` `k) ` `    ``{ ` `       ``Node current = head; ` `       ``Node next = ``null``; ` `       ``Node prev = ``null``; ` `        `  `       ``int` `count = ``0``; ` ` `  `       ``/* Reverse first k nodes of linked list */` `       ``while` `(count < k && current != ``null``)  ` `       ``{ ` `           ``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 now head of input list ` `       ``return` `prev; ` `    ``}                       ` ` `  `                    `  `    ``/* Utility functions */` ` `  `    ``/* Inserts a new Node at front of the list. */` `    ``public` `void` `push(``int` `new_data) ` `    ``{ ` `        ``/* 1 & 2: Allocate the Node & ` `                  ``Put in the data*/` `        ``Node new_node = ``new` `Node(new_data); ` `  `  `        ``/* 3. Make next of new Node as head */` `        ``new_node.next = head; ` `  `  `        ``/* 4. Move the head to point to new Node */` `        ``head = new_node; ` `    ``} ` ` `  `    ``/* Function to print linked list */` `    ``void` `printList() ` `    ``{ ` `        ``Node temp = head; ` `        ``while` `(temp != ``null``) ` `        ``{ ` `           ``System.out.print(temp.data+``" "``); ` `           ``temp = temp.next; ` `        ``}   ` `        ``System.out.println(); ` `    ``} ` ` `  `     ``/* Driver program to test above functions */` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``LinkedList llist = ``new` `LinkedList(); ` `         `  `        ``/* Constructed Linked List is 1->2->3->4->5->6-> ` `           ``7->8->8->9->null */` `        ``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``); ` `         `  `        ``System.out.println(``"Given Linked List"``); ` `        ``llist.printList(); ` `         `  `        ``llist.head = llist.reverse(llist.head, ``3``); ` ` `  `        ``System.out.println(``"Reversed list"``); ` `        ``llist.printList(); ` `    ``} ` `}  ` `/* This code is contributed by Rajat Mishra */`

## Python

 `# 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): ` `        ``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, ` `            ``temp ``=` `temp.``next` ` `  ` `  `# Driver program ` `llist ``=` `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) `

## C#

 `// C# program to reverse a linked list  ` `// in groups of given size  ` `using` `System;  ` ` `  `public` `class` `LinkedList  ` `{  ` `    ``Node head; ``// head of list  ` ` `  `    ``/* Linked list Node*/` `    ``class` `Node  ` `    ``{  ` `        ``public` `int` `data;  ` `        ``public` `Node next;  ` `        ``public` `Node(``int` `d)  ` `        ``{ ` `            ``data = d;  ` `            ``next = ``null``;  ` `        ``}  ` `    ``}  ` ` `  `    ``Node reverse(Node head, ``int` `k)  ` `    ``{  ` `    ``Node current = head;  ` `    ``Node next = ``null``;  ` `    ``Node prev = ``null``;  ` `         `  `    ``int` `count = 0;  ` ` `  `    ``/* Reverse first k nodes of linked list */` `    ``while` `(count < k && current != ``null``)  ` `    ``{  ` `        ``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 now head of input list  ` `    ``return` `prev;  ` `    ``}                  ` ` `  `                     `  `    ``/* Utility functions */` ` `  `    ``/* Inserts a new Node at front of the list. */` `    ``public` `void` `push(``int` `new_data)  ` `    ``{  ` `        ``/* 1 & 2: Allocate the Node &  ` `                ``Put in the data*/` `        ``Node new_node = ``new` `Node(new_data);  ` ` `  `        ``/* 3. Make next of new Node as head */` `        ``new_node.next = head;  ` ` `  `        ``/* 4. Move the head to point to new Node */` `        ``head = new_node;  ` `    ``}  ` ` `  `    ``/* Function to print linked list */` `    ``void` `printList()  ` `    ``{  ` `        ``Node temp = head;  ` `        ``while` `(temp != ``null``)  ` `        ``{  ` `        ``Console.Write(temp.data+``" "``);  ` `        ``temp = temp.next;  ` `        ``}  ` `        ``Console.WriteLine();  ` `    ``}  ` ` `  `    ``/* Driver code*/` `    ``public` `static` `void` `Main(String []args)  ` `    ``{  ` `        ``LinkedList llist = ``new` `LinkedList();  ` `         `  `        ``/* Constructed Linked List is 1->2->3->4->5->6->  ` `        ``7->8->8->9->null */` `        ``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);  ` `         `  `        ``Console.WriteLine(``"Given Linked List"``);  ` `        ``llist.printList();  ` `         `  `        ``llist.head = llist.reverse(llist.head, 3);  ` ` `  `        ``Console.WriteLine(``"Reversed list"``);  ` `        ``llist.printList();  ` `    ``}  ` `}  ` ` `  `// This code is contributed by 29AjayKumar `

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) where n is the number of nodes in the given list.

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

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