Print the alternate nodes of linked list (Iterative Method) in C language

In this problem, program must print the alternates from the given linked list that is leaving one printing other and so on using iterative method.

Iterative method is the one which generally uses loops that are executed till the condition holds value 1 or true.

Let’s say, list contains the nodes 29, 34, 43, 56 and 88 and than the output will be the alternate nodes such as 29, 43 and 88.

Example

Input: 29->34->43->56->88
Output: 29 43 88

The approach is to traverse the entire list till the last node. While, traversing a counter variable can be taken which is incremented to 1 and value is printed when the counter is even or odd depending upon user’s choice. If the user wants it to display from the 0 than the counter with even value is displayed else the counter with odd value is displayed.

The below code shows the c implementation of the algorithm given.

Algorithm

START
   Step 1 -> create node variable of type structure
      Declare int data
      Declare pointer of type node using *next
   Step 2 -> Declare Function void alternate(struct node* head)
      Set int count = 0
      Loop While (head != NULL)
      IF count % 2 = 0
         Print head->data
         Set count++
         Set head = head->next
      End
   Step 3 -> Declare Function void push(struct node** header, int newdata)
      Create newnode using malloc function
      Set newnode->data = newdata
      Set newnode->next = (*header)
      set (*header) = newnode
   step 4 -> In Main()
      create head pointing to first node using struct node* head = NULL
      Call alternate(head)
STOP

Example

#include <stdio.h>
#include <stdlib.h>
//creating structure of a node
struct node {
   int data;
   struct node* next;
};
//function to find and print alternate node
void alternate(struct node* head) {
   int count = 0;
   while (head != NULL) {
      if (count % 2 == 0)
         printf(" %d ", head->data);
      count++;
      head = head->next;
   }
}
//pushing element into the list
void push(struct node** header, int newdata) {
   struct node* newnode =
   (struct node*)malloc(sizeof(struct node));
   newnode->data = newdata;
   newnode->next = (*header);
   (*header) = newnode;
}
int main() {
   printf("alternate nodes are :");
   struct node* head = NULL;
   push(&head, 1); //calling push function to push elements in list
   push(&head, 9);
   push(&head, 10);
   push(&head, 21);
   push(&head, 80);
   alternate(head);
   return 0;
}

Output

If we run above program then it will generate following output.

alternate nodes are : 80 10 1

Sunidhi Bansal

Published on 22-Aug-2019 08:50:52

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