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

  • Difficulty Level : Medium
  • Last Updated : 29 May, 2021

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 

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 an 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)
{
    // base case
    if (!head)
        return NULL;
    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)
{
    if (!head)
        return NULL;
   
    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 code*/
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)
    {
        if(head == null)
          return null;
        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):
       
        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,
            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)
    {
        if(head == null)
          return null;
        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


Javascript




<script>
 
// Javascript program to reverse a
// linked list in groups of
// given size
var head; // head of list
 
    /* Linked list Node */
     class Node {
            constructor(val) {
                this.data = val;
                this.next = null;
            }
        }
 
    function reverse(head , k) {
        if (head == null)
            return null;
        var current = head;
        var next = null;
        var prev = null;
 
        var 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. */
    function push(new_data) {
        /*
         1 & 2: Allocate the Node & Put in the data
         */
        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 */
    function printList() {
        temp = head;
        while (temp != null) {
            document.write(temp.data + " ");
            temp = temp.next;
        }
        document.write("<br/>");
    }
 
    /* Driver program to test above functions */
     
         
        /*
          Constructed Linked List is
          1->2->3->4->5->6-> 7->8->8->9->null
         */
        push(9);
        push(8);
        push(7);
        push(6);
        push(5);
        push(4);
        push(3);
        push(2);
        push(1);
 
        document.write("Given Linked List<br/>");
        printList();
 
        head = reverse(head, 3);
 
        document.write("Reversed list<br/>");
        printList();
 
// This code contributed by gauravrajput1
 
</script>


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(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.
 

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