Print the alternate nodes of linked list (Iterative Method)

Given a linked list, print the alternate nodes of linked list.

Examples:

Input : 1 -> 8 -> 3 -> 10 -> 17 -> 22 -> 29 -> 42
Output : 1 -> 3 -> 17 -> 29
Alternate nodes : 1 -> 3 -> 17 -> 29

Input : 10 -> 17 -> 33 -> 38 -> 73
Output : 10 -> 33 -> 73
Alternate nodes : 10 -> 33 -> 73



Approach :
1. Traverse the whole linked list.
2. Set count = 0.
3. Print node when count is even.
4. Visit the next node.

C

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// CPP code to print Alternate Nodes
#include <stdio.h>
#include <stdlib.h>
  
/* Link list node */
struct Node {
    int data;
    struct Node* next;
};
  
/* Function to get the alternate
   nodes of the linked list */
void printAlternateNode(struct Node* head)
{
    int count = 0;
  
    while (head != NULL) {
  
        // when count is even print the nodes
        if (count % 2 == 0) 
            printf(" %d ", head->data);
  
        // count the nodes
        count++;
  
        // move on the next node.
        head = head->next;
    }
}
  
// Function to push node at head
void push(struct Node** head_ref, int new_data)
{
    struct Node* new_node = 
          (struct Node*)malloc(sizeof(struct Node));
    new_node->data = new_data;
    new_node->next = (*head_ref);
    (*head_ref) = new_node;
}
  
// Driver code
int main()
{
    /* Start with the empty list */
    struct Node* head = NULL;
  
    /* Use push() function to construct
       the below list 8 -> 23 -> 11 -> 29 -> 12 */
    push(&head, 12);
    push(&head, 29);
    push(&head, 11);
    push(&head, 23);
    push(&head, 8);
  
    printAlternateNode(head);
  
    return 0;
}

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Java

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// Java code to print Alternate Nodes 
class GFG
{
      
/* Link list node */
static class Node
    int data; 
    Node next; 
}; 
  
/* Function to get the alternate 
nodes of the linked list */
static void printAlternateNode( Node head) 
    int count = 0
  
    while (head != null)
    
  
        // when count is even print the nodes 
        if (count % 2 == 0
            System.out.printf(" %d ", head.data); 
  
        // count the nodes 
        count++; 
  
        // move on the next node. 
        head = head.next; 
    
  
// Function to push node at head 
static Node push( Node head_ref, int new_data) 
    Node new_node = new Node(); 
    new_node.data = new_data; 
    new_node.next = (head_ref); 
    (head_ref) = new_node;
    return head_ref;
  
// Driver code 
public static void main(String args[]) 
    /* Start with the empty list */
    Node head = null
  
    /* Use push() function to con 
    the below list 8 . 23 . 11 . 29 . 12 */
    head = push(head, 12); 
    head = push(head, 29); 
    head = push(head, 11); 
    head = push(head, 23); 
    head = push(head, 8); 
  
    printAlternateNode(head); 
}
  
// This code is contributed by Arnab Kundu

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Python3

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# Python3 code to print Alternate Nodes
  
# Link list node
class Node : 
      
    def __init__(self, data = None) :
        self.data = data
        self.next = None
      
    # Function to push node at head
    def push(self, data) :
          
        new = Node(data)
        new.next = self
        return new
      
    # Function to get the alternate
    # nodes of the linked list     
    def printAlternateNode(self) :
        head = self
          
        while head and head.next != None :
              
            print(head.data, end = " ")
            head = head.next.next
  
  
# Driver Code        
node = Node()
  
# Use push() function to construct
# the below list 8 -> 23 -> 11 -> 29 -> 12
node = node.push(12)
node = node.push(29)
node = node.push(11)
node = node.push(23)
node = node.push(8)
  
node.printAlternateNode()

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C#

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// C# code to print Alternate Nodes
using System;
  
class GFG 
      
/* Link list node */
public class Node 
    public int data; 
    public Node next; 
}; 
  
/* Function to get the alternate 
nodes of the linked list */
static void printAlternateNode( Node head) 
    int count = 0; 
  
    while (head != null
    
  
        // when count is even print the nodes 
        if (count % 2 == 0) 
            Console.Write(" {0} ", head.data); 
  
        // count the nodes 
        count++; 
  
        // move on the next node. 
        head = head.next; 
    
  
// Function to push node at head 
static Node push( Node head_ref, int new_data) 
    Node new_node = new Node(); 
    new_node.data = new_data; 
    new_node.next = (head_ref); 
    (head_ref) = new_node; 
    return head_ref; 
  
// Driver code 
public static void Main(String []args) 
    /* Start with the empty list */
    Node head = null
  
    /* Use push() function to con 
    the below list 8 . 23 . 11 . 29 . 12 */
    head = push(head, 12); 
    head = push(head, 29); 
    head = push(head, 11); 
    head = push(head, 23); 
    head = push(head, 8); 
  
    printAlternateNode(head); 
  
// This code has been contributed by 29AjayKumar

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Output :

8  11  12

Time Complexity : O(n)
Auxiliary Space : O(1)

Asked in : Govivace



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