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Delete alternate nodes of a Linked List

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  • Difficulty Level : Easy
  • Last Updated : 23 Jun, 2022

Given a Singly Linked List, starting from the second node delete all alternate nodes of it. For example, if the given linked list is 1->2->3->4->5 then your function should convert it to 1->3->5, and if the given linked list is 1->2->3->4 then convert it to 1->3.
 

Method 1 (Iterative) 
Keep track of previous of the node to be deleted. First, change the next link of the previous node and iteratively move to the next node.  
 

C++




// C++ program to remove alternate
// nodes of a linked list
#include <bits/stdc++.h>
using namespace std;
 
/* A linked list node */
class Node
{
    public:
    int data;
    Node *next;
};
 
/* deletes alternate nodes
of a list starting with head */
void deleteAlt(Node *head)
{
    if (head == NULL)
        return;
 
    /* Initialize prev and node to be deleted */
    Node *prev = head;
    Node *node = head->next;
 
    while (prev != NULL && node != NULL)
    {
        /* Change next link of previous node */
        prev->next = node->next;
        delete(node); // delete the node
        /* Update prev and node */
        prev = prev->next;
        if (prev != NULL)
            node = prev->next;
    }
}
 
/* UTILITY FUNCTIONS TO TEST fun1() and fun2() */
/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
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 nodes in a given 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;
 
    /* Using push() to construct below list
    1->2->3->4->5 */
    push(&head, 5);
    push(&head, 4);
    push(&head, 3);
    push(&head, 2);
    push(&head, 1);
 
    cout<<"List before calling deleteAlt() \n";
    printList(head);
 
    deleteAlt(head);
 
    cout<<"\nList after calling deleteAlt() \n";
    printList(head);
 
    return 0;
}
 
// This code is contributed by rathbhupendra

C




// C program to remove alternate nodes of a linked list
#include<stdio.h>
#include<stdlib.h>
 
/* A linked list node */
struct Node
{
    int data;
    struct Node *next;
};
 
/* deletes alternate nodes of a list starting with head */
void deleteAlt(struct Node *head)
{
    if (head == NULL)
        return;
 
    /* Initialize prev and node to be deleted */
    struct Node *prev = head;
    struct Node *node = head->next;
 
    while (prev != NULL && node != NULL)
    {
        /* Change next link of previous node */
        prev->next = node->next;
 
        /* Free memory */
        free(node);
 
        /* Update prev and node */
        prev = prev->next;
        if (prev != NULL)
            node = prev->next;
    }
}
 
/* UTILITY FUNCTIONS TO TEST fun1() and fun2() */
/* Given a reference (pointer to pointer) to the head
  of a list and an int, push a new node on the front
  of the list. */
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 nodes in a given linked list */
void printList(struct Node *node)
{
    while (node != NULL)
    {
        printf("%d ", node->data);
        node = node->next;
    }
}
 
/* Driver program to test above functions */
int main()
{
    /* Start with the empty list */
    struct Node* head = NULL;
 
    /* Using push() to construct below list
      1->2->3->4->5  */
    push(&head, 5);
    push(&head, 4);
    push(&head, 3);
    push(&head, 2);
    push(&head, 1);
 
    printf("\nList before calling deleteAlt() \n");
    printList(head);
 
    deleteAlt(head);
 
    printf("\nList after calling deleteAlt() \n");
    printList(head);
 
    return 0;
}

Java




// Java program to delete alternate nodes of a linked list
class LinkedList {
    Node head; // head of list
 
    /* Linked list Node*/
    class Node {
        int data;
        Node next;
        Node(int d)
        {
            data = d;
            next = null;
        }
    }
 
    void deleteAlt()
    {
        if (head == null)
            return;
 
        Node node = head;
 
        while (node != null && node.next != null) {
            /* Change next link of next node */
            node.next = node.next.next;
 
            /*Update ref node to new alternate node */
            node = node.next;
        }
    }
 
    /* 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->null */
        llist.push(5);
        llist.push(4);
        llist.push(3);
        llist.push(2);
        llist.push(1);
 
        System.out.println(
            "Linked List before calling deleteAlt() ");
        llist.printList();
 
        llist.deleteAlt();
 
        System.out.println(
            "Linked List after calling deleteAlt() ");
        llist.printList();
    }
}
/* This code is contributed by Rajat Mishra */

Python3




# Python3 program to remove alternate
# nodes of a linked list
import math
 
# A linked list node
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None
         
# deletes alternate nodes
# of a list starting with head
def deleteAlt(head):
    if (head == None):
        return
 
    # Initialize prev and node to be deleted
    prev = head
    now = head.next
 
    while (prev != None and now != None):
         
        # Change next link of previous node
        prev.next = now.next
 
        # Free memory
        now = None
 
        # Update prev and node
        prev = prev.next
        if (prev != None):
            now = prev.next
     
# UTILITY FUNCTIONS TO TEST fun1() and fun2()
# Given a reference (pointer to pointer) to the head
# of a list and an , push a new node on the front
# of the list.
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 point to the new node
    head_ref = new_node
    return head_ref
 
# Function to print nodes in a given linked list
def printList(node):
    while (node != None):
        print(node.data, end = " ")
        node = node.next
     
# Driver code
if __name__=='__main__':
     
    # Start with the empty list
    head = None
 
    # Using head=push() to construct below list
    # 1.2.3.4.5
    head = push(head, 5)
    head = push(head, 4)
    head = push(head, 3)
    head = push(head, 2)
    head = push(head, 1)
 
    print("List before calling deleteAlt() ")
    printList(head)
 
    deleteAlt(head)
 
    print("\nList after calling deleteAlt() ")
    printList(head)
 
# This code is contributed by Srathore

C#




// C# program to delete alternate
// nodes of a linked list
using System;
 
public class LinkedList
{
    Node head; // head of list
 
    /* Linked list Node*/
    public class Node
    {
        public int data;
        public Node next;
        public Node(int d)
        {
            data = d; next = null;
             
        }
    }
 
    void deleteAlt()
    {
        if (head == null)
            return;
 
        Node prev = head;
        Node now = head.next;
 
        while (prev != null && now != null)
        {        
            /* Change next link of previous node */
            prev.next = now.next;
 
            /* Free node */
            now = null;
 
            /*Update prev and now */
            prev = prev.next;
            if (prev != null)
                now = prev.next;
        }
    }                
 
                     
    /* 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->null */
        llist.push(5);
        llist.push(4);
        llist.push(3);
        llist.push(2);
        llist.push(1);
         
        Console.WriteLine("Linked List before" +
                            "calling deleteAlt() ");
        llist.printList();
         
        llist.deleteAlt();
         
        Console.WriteLine("Linked List after" +
                            "calling deleteAlt() ");
        llist.printList();
    }
}
 
// This code has been contributed
// by 29AjayKumar

Javascript




<script>
 
// Javascript program to delete alternate
// nodes of a linked list
var head; // head of list
 
    /* Linked list Node */
     class Node {
            constructor(val) {
                this.data = val;
                this.next = null;
            }
        }
      
    function deleteAlt() {
        if (head == null)
            return;
 
        var prev = head;
        var now = head.next;
 
        while (prev != null && now != null) {
            /* Change next link of previous node */
            prev.next = now.next;
 
            /* Free node */
            now = null;
 
            /* Update prev and now */
            prev = prev.next;
            if (prev != null)
                now = prev.next;
        }
    }
 
    /* Utility functions */
 
    /* Inserts a new Node at front of the list. */
    function push(new_data) {
        /*
         * 1 & 2: Allocate the Node & Put in the data
         */
        var 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() {
        var 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->null */
        push(5);
        push(4);
        push(3);
        push(2);
        push(1);
 
        document.write(
        "Linked List before calling deleteAlt() <br/>"
        );
        printList();
 
        deleteAlt();
 
        document.write(
        "Linked List after calling deleteAlt()<br/> "
        );
        printList();
 
// This code contributed by gauravrajput1
 
</script>

Output

List before calling deleteAlt() 
1 2 3 4 5 
List after calling deleteAlt() 
1 3 5 

Time Complexity: O(n) 

where n is the number of nodes in the given Linked List.

Auxiliary Space: O(1)

As constant extra space is used.

Method 2 (Recursive) 
Recursive code uses the same approach as method 1. The recursive code is simple and short but causes O(n) recursive function calls for a linked list of size n. 
 

C++




/* deletes alternate nodes of a list starting with head */
void deleteAlt(Node *head)
{
    if (head == NULL)
        return;
 
    Node *node = head->next;
 
    if (node == NULL)
        return;
 
    /* Change the next link of head */
    head->next = node->next;
 
    /* free memory allocated for node */
    free(node);
 
    /* Recursively call for the new next of head */
    deleteAlt(head->next);
}
 
// This code is contributed by rathbhupendra

C




/* deletes alternate nodes of a list starting with head */
void deleteAlt(struct Node *head)
{
    if (head == NULL)
        return;
 
    struct Node *node = head->next;
 
    if (node == NULL)
        return;
 
    /* Change the next link of head */
    head->next = node->next;
 
    /* free memory allocated for node */
    free(node);
 
    /* Recursively call for the new next of head */
    deleteAlt(head->next);
}

Java




/* deletes alternate nodes of a list
starting with head */
static Node deleteAlt(Node head)
{
    if (head == null)
        return;
 
    Node node = head.next;
 
    if (node == null)
        return;
 
    /* Change the next link of head */
    head.next = node.next;
 
 
    /* Recursively call for the new next of head */
    head.next = deleteAlt(head.next);
}
 
// This code is contributed by Arnab Kundu

Python3




# deletes alternate nodes of a list starting with head
def deleteAlt(head):
    if (head == None):
        return
 
    node = head.next
 
    if (node == None):
        return
 
    # Change the next link of head
    head.next = node.next
 
    # free memory allocated for node
    #free(node)
 
    # Recursively call for the new next of head
    deleteAlt(head.next)
 
# This code is contributed by Srathore

C#




/* deletes alternate nodes of a list
starting with head */
static Node deleteAlt(Node head)
{
    if (head == null)
        return;
 
    Node node = head.next;
 
    if (node == null)
        return;
 
    /* Change the next link of head */
    head.next = node.next;
 
 
    /* Recursively call for the new next of head */
    head.next = deleteAlt(head.next);
}
 
// This code is contributed by Arnab Kundu

Javascript




<script>
/* deletes alternate nodes of a list
starting with head */
function deleteAlt(head)
{
    if (head == null)
        return;
 
    var node = head.next;
 
    if (node == null)
        return;
 
    /* Change the next link of head */
    head.next = node.next;    /* Recursively call for the new next of head */
    head.next = deleteAlt(head.next);
}
 
 
// This code contributed by aashish1995
</script>

Time Complexity: O(n)

Auxiliary Space: O(1)

As this is a tail recursive function no function call stack is required thus the extra space used is constant.
 

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


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