Explain the deletion of element in linked list

Linked lists use dynamic memory allocation i.e. they grow and shrink accordingly. They are defined as a collection of nodes. Here, nodes have two parts, which are data and link. The representation of data, link and linked lists is given below −

Operations on linked lists

There are three types of operations on linked lists in C language, which are as follows −

• Insertion
• Deletion
• Traversing

Deletion

Consider an example given below −

Delete node 2

Delete node 1

Delete node 3

Program

Following is the C program for deletion of the elements in linked lists −

 Live Demo

#include <stdio.h>
#include <stdlib.h>
struct Node{
   int data;
   struct Node *next;
};
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;
}
void deleteNode(struct Node **head_ref, int position){
   //if list is empty
   if (*head_ref == NULL)
      return;
   struct Node* temp = *head_ref;
   if (position == 0){
      *head_ref = temp->next;
      free(temp);
      return;
   }
   for (int i=0; temp!=NULL && i<position-1; i++)
      temp = temp->next;
   if (temp == NULL || temp->next == NULL)
      return;
   struct Node *next = temp->next->next;
   free(temp->next); // Free memory
   temp->next = next;
}
void printList(struct Node *node){
   while (node != NULL){
      printf(" %d ", node->data);
      node = node->next;
   }
}
int main(){
   struct Node* head = NULL;
   push(&head, 7);
   push(&head, 1);
   push(&head, 3);
   push(&head, 2);
   push(&head, 8);
   puts("Created List: ");
   printList(head);
   deleteNode(&head, 3);
   puts("
 List after Deletion at position 3: ");
   printList(head);
   return 0;
}

Output

When the above program is executed, it produces the following result −

Created List:
8 2 3 1 7
List after Deletion at position 3:
8 2 3 7