Reverse alternate K nodes in a Singly Linked List

Given a linked list, write a function to reverse every alternate k nodes (where k is an input to the function) in an efficient way. Give the complexity of your algorithm.

Example:

Inputs:   1->2->3->4->5->6->7->8->9->NULL and k = 3
Output:   3->2->1->4->5->6->9->8->7->NULL.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1 (Process 2k nodes and recursively call for rest of the list)
This method is basically an extension of the method discussed in this post.

kAltReverse(struct node *head, int k)
  1)  Reverse first k nodes.
  2)  In the modified list head points to the kth node.  So change next 
       of head to (k+1)th node
  3)  Move the current pointer to skip next k nodes.
  4)  Call the kAltReverse() recursively for rest of the n - 2k nodes.
  5)  Return new head of the list.

C++

// C++ program to reverse alternate 
// k nodes in a linked list 
#include <bits/stdc++.h> 
using namespace std; 
  
/* Link list node */
class Node  
{  
    public: 
    int data;  
    Node* next;  
};  
  
/* Reverses alternate k nodes and  
returns the pointer to the new head node */
Node *kAltReverse(Node *head, int k)  
{  
    Node* current = head;  
    Node* next;  
    Node* prev = NULL;  
    int count = 0;  
  
    /*1) reverse first k nodes of the linked list */
    while (current != NULL && count < k)  
    {  
    next = current->next;  
    current->next = prev;  
    prev = current;  
    current = next;  
    count++;  
    }  
      
    /* 2) Now head points to the kth node.  
    So change next  of head to (k+1)th node*/
    if(head != NULL)  
    head->next = current;  
  
    /* 3) We do not want to reverse next k  
       nodes. So move the current  
        pointer to skip next k nodes */
    count = 0;  
    while(count < k-1 && current != NULL )  
    {  
    current = current->next;  
    count++;  
    }  
  
    /* 4) Recursively call for the list  
    starting from current->next. And make 
    rest of the list as next of first node */
    if(current != NULL)  
    current->next = kAltReverse(current->next, k);  
  
    /* 5) 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)  
{  
    int count = 0;  
    while(node != NULL)  
    {  
        cout<<node->data<<" ";  
        node = node->next;  
        count++;  
    }  
}  
  
/* Driver code*/
int main(void)  
{  
    /* Start with the empty list */
    Node* head = NULL;  
    int i;  
      
    // create a list 1->2->3->4->5...... ->20  
    for(i = 20; i > 0; i--)  
    push(&head, i);  
  
    cout<<"Given linked list \n";  
    printList(head);  
    head = kAltReverse(head, 3);  
  
    cout<<"\n Modified Linked list \n";  
    printList(head);  
    return(0);  
}  
  
  
// This code is contributed by rathbhupendra 

C

#include<stdio.h> 
#include<stdlib.h> 
  
/* Link list node */
struct Node 
{ 
    int data; 
    struct Node* next; 
}; 
  
/* Reverses alternate k nodes and 
   returns the pointer to the new head node */
struct Node *kAltReverse(struct Node *head, int k) 
{ 
    struct Node* current = head; 
    struct Node* next; 
    struct Node* prev = NULL; 
    int count = 0;    
  
    /*1) reverse first k nodes of the linked list */
    while (current != NULL && count < k) 
    { 
       next  = current->next; 
       current->next = prev; 
       prev = current; 
       current = next; 
       count++; 
    } 
    
    /* 2) Now head points to the kth node.  So change next  
       of head to (k+1)th node*/ 
    if(head != NULL) 
      head->next = current;    
  
    /* 3) We do not want to reverse next k nodes. So move the current  
        pointer to skip next k nodes */
    count = 0; 
    while(count < k-1 && current != NULL ) 
    { 
      current = current->next; 
      count++; 
    } 
  
    /* 4) Recursively call for the list starting from current->next. 
       And make rest of the list as next of first node */
    if(current !=  NULL) 
       current->next = kAltReverse(current->next, k);  
  
    /* 5) 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) 
{ 
    int count = 0; 
    while(node != NULL) 
    { 
        printf("%d  ", node->data); 
        node = node->next; 
        count++; 
    } 
}     
  
/* Driver program to test above function*/
int main(void) 
{ 
    /* Start with the empty list */
    struct Node* head = NULL; 
    int i; 
    // create a list 1->2->3->4->5...... ->20 
    for(i = 20; i > 0; i--) 
      push(&head, i); 
  
     printf("\n Given linked list \n"); 
     printList(head); 
     head = kAltReverse(head, 3); 
  
     printf("\n Modified Linked list \n"); 
     printList(head); 
  
     getchar(); 
     return(0); 
} 

Java

// Java program to reverse alternate k nodes in a linked list 
  
class LinkedList { 
  
    static Node head; 
  
    class Node { 
  
        int data; 
        Node next; 
  
        Node(int d) { 
            data = d; 
            next = null; 
        } 
    } 
  
    /* Reverses alternate k nodes and 
     returns the pointer to the new head node */
    Node kAltReverse(Node node, int k) { 
        Node current = node; 
        Node next = null, prev = null; 
        int count = 0; 
  
        /*1) reverse first k nodes of the linked list */
        while (current != null && count < k) { 
            next = current.next; 
            current.next = prev; 
            prev = current; 
            current = next; 
            count++; 
        } 
  
        /* 2) Now head points to the kth node.  So change next  
         of head to (k+1)th node*/
        if (node != null) { 
            node.next = current; 
        } 
  
        /* 3) We do not want to reverse next k nodes. So move the current  
         pointer to skip next k nodes */
        count = 0; 
        while (count < k - 1 && current != null) { 
            current = current.next; 
            count++; 
        } 
  
        /* 4) Recursively call for the list starting from current->next. 
         And make rest of the list as next of first node */
        if (current != null) { 
            current.next = kAltReverse(current.next, k); 
        } 
  
        /* 5) prev is new head of the input list */
        return prev; 
    } 
  
    void printList(Node node) { 
        while (node != null) { 
            System.out.print(node.data + " "); 
            node = node.next; 
        } 
    } 
  
    void push(int newdata) { 
        Node mynode = new Node(newdata); 
        mynode.next = head; 
        head = mynode; 
    } 
  
    public static void main(String[] args) { 
        LinkedList list = new LinkedList(); 
  
        // Creating the linkedlist 
        for (int i = 20; i > 0; i--) { 
            list.push(i); 
        } 
        System.out.println("Given Linked List :"); 
        list.printList(head); 
        head = list.kAltReverse(head, 3); 
        System.out.println(""); 
        System.out.println("Modified Linked List :"); 
        list.printList(head); 
  
    } 
} 
  
// This code has been contributed by Mayank Jaiswal 

Python3

# Python3 program to reverse alternate 
# k nodes in a linked list 
import math 
  
# Link list node  
class Node:  
    def __init__(self, data):  
        self.data = data  
        self.next = None
  
# Reverses alternate k nodes and  
#returns the poer to the new head node  
def kAltReverse(head, k) :  
    current = head  
    next = None
    prev = None
    count = 0
  
    #1) reverse first k nodes of the linked list  
    while (current != None and count < k) :  
        next = current.next
        current.next = prev  
        prev = current  
        current = next
        count = count + 1; 
      
    # 2) Now head pos to the kth node.  
    # So change next of head to (k+1)th node 
    if(head != None):  
        head.next = current  
  
    # 3) We do not want to reverse next k  
    # nodes. So move the current  
    # poer to skip next k nodes  
    count = 0
    while(count < k - 1 and current != None ):  
        current = current.next
        count = count + 1
      
    # 4) Recursively call for the list  
    # starting from current.next. And make 
    # rest of the list as next of first node  
    if(current != None):  
        current.next = kAltReverse(current.next, k)  
  
    # 5) prev is new head of the input list  
    return prev  
  
# UTILITY FUNCTIONS  
# Function to push a node  
def push(head_ref, new_data):  
      
    # allocate node  
    new_node = Node(new_data) 
  
    # 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 po to the new node  
    head_ref = new_node 
      
    return head_ref 
  
# Function to print linked list  
def prList(node):  
    count = 0
    while(node != None):  
        print(node.data, end = " ")  
        node = node.next
        count = count + 1
      
# Driver code 
if __name__=='__main__': 
      
    # Start with the empty list  
    head = None
  
    # create a list 1.2.3.4.5...... .20  
    for i in range(20, 0, -1): 
        head = push(head, i)  
          
    print("Given linked list ")  
    prList(head)  
    head = kAltReverse(head, 3)  
  
    print("\nModified Linked list")  
    prList(head)  
      
# This code is contributed by Srathore 

C#

// C# program to reverse alternate 
// k nodes in a linked list  
using System; 
class LinkedList  
{  
  
    static Node head;  
  
    public class Node  
    {  
  
        public int data;  
        public Node next;  
  
        public Node(int d) 
        {  
            data = d;  
            next = null;  
        }  
    }  
  
    /* Reverses alternate k nodes and  
    returns the pointer to the new head node */
    Node kAltReverse(Node node, int k)  
    {  
        Node current = node;  
        Node next = null, prev = null;  
        int count = 0;  
  
        /*1) reverse first k nodes of the linked list */
        while (current != null && count < k) 
        {  
            next = current.next;  
            current.next = prev;  
            prev = current;  
            current = next;  
            count++;  
        }  
  
        /* 2) Now head points to the kth  
        node. So change next  
        of head to (k+1)th node*/
        if (node != null)  
        {  
            node.next = current;  
        }  
  
        /* 3) We do not want to reverse  
        next k nodes. So move the current  
        pointer to skip next k nodes */
        count = 0;  
        while (count < k - 1 && current != null)  
        {  
            current = current.next;  
            count++;  
        }  
  
        /* 4) Recursively call for the  
        list starting from current->next.  
        And make rest of the list as  
        next of first node */
        if (current != null) 
        {  
            current.next = kAltReverse(current.next, k);  
        }  
  
        /* 5) prev is new head of the input list */
        return prev;  
    }  
  
    void printList(Node node)  
    {  
        while (node != null)  
        {  
            Console.Write(node.data + " ");  
            node = node.next;  
        }  
    }  
  
    void push(int newdata) 
    {  
        Node mynode = new Node(newdata);  
        mynode.next = head;  
        head = mynode;  
    }  
  
    // Driver code 
    public static void Main(String []args) 
    {  
        LinkedList list = new LinkedList();  
  
        // Creating the linkedlist  
        for (int i = 20; i > 0; i--)  
        {  
            list.push(i);  
        }  
        Console.WriteLine("Given Linked List :");  
        list.printList(head);  
        head = list.kAltReverse(head, 3);  
        Console.WriteLine("");  
        Console.WriteLine("Modified Linked List :");  
        list.printList(head);  
    }  
}  
  
// This code has been contributed by Arnab Kundu 

Output:

Given linked list
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Modified Linked list
3 2 1 4 5 6 9 8 7 10 11 12 15 14 13 16 17 18 20 19

Time Complexity: O(n)

Method 2 (Process k nodes and recursively call for rest of the list)
The method 1 reverses the first k node and then moves the pointer to k nodes ahead. So method 1 uses two while loops and processes 2k nodes in one recursive call.
This method processes only k nodes in a recursive call. It uses a third bool parameter b which decides whether to reverse the k elements or simply move the pointer.

_kAltReverse(struct node *head, int k, bool b)
  1)  If b is true, then reverse first k nodes.
  2)  If b is false, then move the pointer k nodes ahead.
  3)  Call the kAltReverse() recursively for rest of the n - k nodes and link 
       rest of the modified list with end of first k nodes. 
  4)  Return new head of the list.

C++

#include <bits/stdc++.h> 
using namespace std; 
  
/* Link list node */
class node  
{  
    public: 
    int data;  
    node* next;  
};  
  
/* Helper function for kAltReverse() */
node * _kAltReverse(node *node, int k, bool b);  
  
/* Alternatively reverses the given linked list  
in groups of given size k. */
node *kAltReverse(node *head, int k)  
{  
    return _kAltReverse(head, k, true);  
}  
  
/* Helper function for kAltReverse().   
It reverses k nodes of the list only if  
the third parameter b is passed as true,  
otherwise moves the pointer k nodes ahead 
and recursively calls iteself */
node * _kAltReverse(node *Node, int k, bool b)  
{  
    if(Node == NULL)  
        return NULL;  
      
    int count = 1;  
    node *prev = NULL;  
    node *current = Node;  
    node *next;  
      
    /* The loop serves two purposes  
        1) If b is true,  
           then it reverses the k nodes  
        2) If b is false,  
           then it moves the current pointer */
    while(current != NULL && count <= k)  
    {  
        next = current->next;  
      
        /* Reverse the nodes only if b is true*/
        if(b == true)  
            current->next = prev;  
                  
        prev = current;  
        current = next;  
        count++;  
    }  
          
    /* 3) If b is true, then node is the kth node.  
        So attach rest of the list after node.  
        4) After attaching, return the new head */
    if(b == true)  
    {  
        Node->next = _kAltReverse(current, k, !b);  
        return prev;          
    }  
          
    /* If b is not true, then attach  
    rest of the list after prev.  
    So attach rest of the list after prev */
    else
    {  
        prev->next = _kAltReverse(current, k, !b);  
        return Node;      
    }  
}  
  
  
/* 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)  
{  
    int count = 0;  
    while(node != NULL)  
    {  
        cout << node->data << " ";  
        node = node->next;  
        count++;  
    }  
}  
  
// Driver Code 
int main(void)  
{  
    /* Start with the empty list */
    node* head = NULL;  
    int i;  
  
    // create a list 1->2->3->4->5...... ->20  
    for(i = 20; i > 0; i--)  
    push(&head, i);  
  
    cout << "Given linked list \n";  
    printList(head);  
    head = kAltReverse(head, 3);  
  
    cout << "\nModified Linked list \n";  
    printList(head);  
    return(0);  
}  
  
// This code is contributed by rathbhupendra 

C

#include<stdio.h> 
#include<stdlib.h> 
  
/* Link list node */
struct node 
{ 
    int data; 
    struct node* next; 
}; 
  
/* Helper function for kAltReverse() */
struct node * _kAltReverse(struct node *node, int k, bool b); 
  
/* Alternatively reverses the given linked list in groups of  
   given size k. */
struct node *kAltReverse(struct node *head, int k) 
{ 
  return _kAltReverse(head, k, true); 
} 
   
/*  Helper function for kAltReverse().  It reverses k nodes of the list only if 
    the third parameter b is passed as true, otherwise moves the pointer k  
    nodes ahead and recursively calls iteself  */ 
struct node * _kAltReverse(struct node *node, int k, bool b) 
{ 
   if(node == NULL) 
       return NULL; 
  
   int count = 1; 
   struct node *prev = NULL; 
   struct node  *current = node; 
   struct node *next; 
   
   /* The loop serves two purposes 
      1) If b is true, then it reverses the k nodes  
      2) If b is false, then it moves the current pointer */
   while(current != NULL && count <= k) 
   { 
       next = current->next; 
  
       /* Reverse the nodes only if b is true*/
       if(b == true) 
          current->next = prev; 
              
       prev = current; 
       current = next; 
       count++; 
   } 
     
   /* 3) If b is true, then node is the kth node.  
       So attach rest of the list after node.  
     4) After attaching, return the new head */
   if(b == true) 
   { 
        node->next = _kAltReverse(current,k,!b); 
        return prev;         
   } 
     
   /* If b is not true, then attach rest of the list after prev.  
     So attach rest of the list after prev */    
   else
   { 
        prev->next = _kAltReverse(current, k, !b); 
        return node;        
   } 
} 
   
  
/* 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) 
{ 
    int count = 0; 
    while(node != NULL) 
    { 
        printf("%d  ", node->data); 
        node = node->next; 
        count++; 
    } 
} 
   
/* Driver program to test above function*/
int main(void) 
{ 
    /* Start with the empty list */
    struct node* head = NULL; 
    int i; 
   
    // create a list 1->2->3->4->5...... ->20 
    for(i = 20; i > 0; i--) 
      push(&head, i); 
   
    printf("\n Given linked list \n"); 
    printList(head); 
    head = kAltReverse(head, 3); 
   
    printf("\n Modified Linked list \n"); 
    printList(head); 
   
    getchar(); 
    return(0); 
} 

Java

// Java program to reverse alternate k nodes in a linked list 
  
class LinkedList { 
  
    static Node head; 
  
    class Node { 
  
        int data; 
        Node next; 
  
        Node(int d) { 
            data = d; 
            next = null; 
        } 
    } 
  
    /* Alternatively reverses the given linked list in groups of  
     given size k. */
    Node kAltReverse(Node head, int k) { 
        return _kAltReverse(head, k, true); 
    } 
  
    /*  Helper function for kAltReverse().  It reverses k nodes of the list only if 
     the third parameter b is passed as true, otherwise moves the pointer k  
     nodes ahead and recursively calls iteself  */
    Node _kAltReverse(Node node, int k, boolean b) { 
        if (node == null) { 
            return null; 
        } 
  
        int count = 1; 
        Node prev = null; 
        Node current = node; 
        Node next = null; 
  
        /* The loop serves two purposes 
         1) If b is true, then it reverses the k nodes  
         2) If b is false, then it moves the current pointer */
        while (current != null && count <= k) { 
            next = current.next; 
  
            /* Reverse the nodes only if b is true*/
            if (b == true) { 
                current.next = prev; 
            } 
  
            prev = current; 
            current = next; 
            count++; 
        } 
  
        /* 3) If b is true, then node is the kth node.  
         So attach rest of the list after node.  
         4) After attaching, return the new head */
        if (b == true) { 
            node.next = _kAltReverse(current, k, !b); 
            return prev; 
        } /* If b is not true, then attach rest of the list after prev.  
         So attach rest of the list after prev */ else { 
            prev.next = _kAltReverse(current, k, !b); 
            return node; 
        } 
    } 
  
    void printList(Node node) { 
        while (node != null) { 
            System.out.print(node.data + " "); 
            node = node.next; 
        } 
    } 
  
    void push(int newdata) { 
        Node mynode = new Node(newdata); 
        mynode.next = head; 
        head = mynode; 
    } 
  
    public static void main(String[] args) { 
        LinkedList list = new LinkedList(); 
  
        // Creating the linkedlist 
        for (int i = 20; i > 0; i--) { 
            list.push(i); 
        } 
        System.out.println("Given Linked List :"); 
        list.printList(head); 
        head = list.kAltReverse(head, 3); 
        System.out.println(""); 
        System.out.println("Modified Linked List :"); 
        list.printList(head); 
  
    } 
} 
  
// This code has been contributed by Mayank Jaiswal 

Python

# Python code for above algorithm 
  
# Link list node  
class node:  
      
    def __init__(self, data):  
        self.data = data  
        self.next = next
          
# function to insert a node at the 
# beginning of the linked list 
def push(head_ref, new_data): 
  
    # allocate node  
    new_node = node(0) 
  
    # put in the data  
    new_node.data = new_data 
  
    # link the old list to the new node  
    new_node.next = (head_ref) 
  
    # move the head to point to the new node  
    (head_ref) = new_node 
      
    return head_ref 
      
""" Alternatively reverses the given linked list  
in groups of given size k. """
def kAltReverse(head, k) : 
  
    return _kAltReverse(head, k, True)  
  
""" Helper function for kAltReverse().  
It reverses k nodes of the list only if  
the third parameter b is passed as True,  
otherwise moves the pointer k nodes ahead 
and recursively calls iteself """
def _kAltReverse(Node, k, b) : 
  
    if(Node == None) : 
        return None
      
    count = 1
    prev = None
    current = Node  
    next = None
      
    """ The loop serves two purposes  
        1) If b is True,  
        then it reverses the k nodes  
        2) If b is false,  
        then it moves the current pointer """
    while(current != None and count <= k) : 
      
        next = current.next
      
        """ Reverse the nodes only if b is True"""
        if(b == True) : 
            current.next = prev  
                  
        prev = current  
        current = next
        count = count + 1
      
          
    """ 3) If b is True, then node is the kth node.  
        So attach rest of the list after node.  
        4) After attaching, return the new head """
    if(b == True) : 
      
        Node.next = _kAltReverse(current, k, not b)  
        return prev          
      
    else : 
        """ If b is not True, then attach  
        rest of the list after prev.  
        So attach rest of the list after prev """
        prev.next = _kAltReverse(current, k, not b)  
        return Node      
      
""" Function to print linked list """
def printList(node) : 
  
    count = 0
    while(node != None) : 
      
        print( node.data, end = " ")  
        node = node.next
        count = count + 1
  
# Driver Code 
  
""" Start with the empty list """
head = None
i = 20
  
# create a list 1->2->3->4->5...... ->20  
while(i > 0 ):  
    head = push(head, i)  
    i = i - 1
  
print( "Given linked list ")  
printList(head)  
head = kAltReverse(head, 3)  
  
print( "\nModified Linked list ")  
printList(head)  
  
# This code is contributed by Arnab Kundu 

C#

// C# Program for converting  
// singly linked list into  
// circular linked list.  
using System; 
  
public class LinkedList 
{  
  
    static Node head;  
  
    public class Node  
    {  
  
        public int data;  
        public Node next;  
  
        public Node(int d)  
        {  
            data = d;  
            next = null;  
        }  
    }  
  
    /* Reverses alternate k nodes and  
    returns the pointer to the new head node */
    Node kAltReverse(Node node, int k)  
    {  
        Node current = node;  
        Node next = null, prev = null;  
        int count = 0;  
  
        /*1) reverse first k nodes of the linked list */
        while (current != null && count < k)  
        {  
            next = current.next;  
            current.next = prev;  
            prev = current;  
            current = next;  
            count++;  
        }  
  
        /* 2) Now head points to the kth node.  
        So change next of head to (k+1)th node*/
        if (node != null) 
        {  
            node.next = current;  
        }  
  
        /* 3) We do not want to reverse next  
        k nodes. So move the current  
        pointer to skip next k nodes */
        count = 0;  
        while (count < k - 1 && current != null) 
        {  
            current = current.next;  
            count++;  
        }  
  
        /* 4) Recursively call for the list  
        starting from current->next. And make 
         rest of the list as next of first node */
        if (current != null)  
        {  
            current.next = kAltReverse(current.next, k);  
        }  
  
        /* 5) prev is new head of the input list */
        return prev;  
    }  
  
    void printList(Node node)  
    {  
        while (node != null)  
        {  
            Console.Write(node.data + " ");  
            node = node.next;  
        }  
    }  
  
    void push(int newdata) 
    {  
        Node mynode = new Node(newdata);  
        mynode.next = head;  
        head = mynode;  
    }  
  
    public static void Main(String[] args)  
    {  
        LinkedList list = new LinkedList();  
  
        // Creating the linkedlist  
        for (int i = 20; i > 0; i--) 
        {  
            list.push(i);  
        }  
        Console.WriteLine("Given Linked List :");  
        list.printList(head);  
        head = list.kAltReverse(head, 3);  
        Console.WriteLine("");  
        Console.WriteLine("Modified Linked List :");  
        list.printList(head);  
    }  
}  
  
// This code is contributed 29AjayKumar 

Output:

Given linked list
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Modified Linked list
3 2 1 4 5 6 9 8 7 10 11 12 15 14 13 16 17 18 20 19

Time Complexity: O(n)
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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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C

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