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

3.3

Given a linked list, write a function to reverse every k nodes (where k is an input to the function).

Example:
Inputs:  1->2->3->4->5->6->7->8->NULL and k = 3 
Output:  3->2->1->6->5->4->8->7->NULL. 

Inputs:   1->2->3->4->5->6->7->8->NULL and k = 5
Output:  5->4->3->2->1->8->7->6->NULL. 

Algorithm: reverse(head, k)
1) 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.
2) head->next = reverse(next, k) /* Recursively call for rest of the list and link the two sub-lists */
3) return prev /* prev becomes the new head of the list (see the diagrams of iterative method of this post) */

C/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;
    }
}    

/* Drier 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();
    }

     /* Drier 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)


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.

Asked in: Adobe, Microsoft, Paytm, Snapdeal

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

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