Reverse a Linked List in groups of given size

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.

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 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 <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)  
{  
    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<<node->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<stdio.h> 
#include<stdlib.h> 
  
/* 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

Complexity Analysis:

Time Complexity: O(n).
Traversal of list is done only once and it has ‘n’ elements.
Auxiliary Space: O(1).
No use of data structure for storing values.

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

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My Personal Notes

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

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

C++

C

Java

Python

C#

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