Delete alternate nodes of a Linked List

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 previous node and then free the memory allocated for the node.

C/C++

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// 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;
    }
}
  
/* Drier 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;
}

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Java

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// 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 prev = head;
       Node now = head.next;
  
       while (prev != null && now != null
       {           
           /* Change next link of previus 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)
        {
           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->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 */

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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.

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

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/* 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);
}

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Time Complexity: O(n)

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



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