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C++ Program To Delete Alternate Nodes Of A Linked List

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  • Last Updated : 31 Aug, 2022
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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;
 
        // 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 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() ";
    printList(head);
 
    deleteAlt(head);
 
    cout << "List after calling deleteAlt() ";
    printList(head);
 
    return 0;
}
// This code is contributed by rathbhupendra


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) because it is using constant space

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


Time Complexity: O(n)

Auxiliary space: O(n) for call stack because using recursion

Please refer complete article on Delete alternate nodes of a Linked List for more details!


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