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C Program To Delete Alternate Nodes Of A Linked List

  • Last Updated : 16 Dec, 2021

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<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 code
int main()
{
    // Start with the empty list 
    struct 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);
  
    printf("\nList before calling deleteAlt() \n");
    printList(head);
  
    deleteAlt(head);
  
    printf("\nList after calling deleteAlt() \n");
    printList(head);
    return 0;
}

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 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(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);
}

Time Complexity: O(n)

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


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