Explain insertion of elements in linked list using C language

C Server Side Programming Programming

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

Insertion

Consider an example, wherein we insert node 5 in between node 2 and node 3.

Now, insert node 5 at the beginning.

Insert node 5 at the end.

Note:

We cannot insert node 5 before node 2 as the nodes are not named.
We can insert node 5 before 2, if its position is given.

Program

Following is the C program for inserting an element in linked list −

Live Demo

#include <stdio.h>
#include <stdlib.h>
struct node{
   int val;
   struct node *next;
};
void print_list(struct node *head){
   printf("H->");
   while(head){
      printf("%d->", head->val);
      head = head->next;
   }
   printf("……

");
}
void insert_front(struct node **head, int value){
   struct node * new_node = NULL;
   new_node = (struct node *)malloc(sizeof(struct node));
   if (new_node == NULL){
      printf(" Out of memory");
   }
   new_node->val = value;
   new_node->next = *head;
   *head = new_node;
}
void insert_end(struct node **head, int value){
   struct node * new_node = NULL;
   struct node * last = NULL;
   new_node = (struct node *)malloc(sizeof(struct node));
   if (new_node == NULL){
      printf(" Out of memory");
   }
   new_node->val = value;
   new_node->next = NULL;
   if( *head == NULL){
      *head = new_node;
      return;
   }
   last = *head;
   while(last->next) last = last->next;
   last->next = new_node;
}
void insert_after(struct node *head, int value, int after){
   struct node * new_node = NULL;
   struct node *tmp = head;
   while(tmp) {
      if(tmp->val == after) { /*found the node*/
         new_node = (struct node *)malloc(sizeof(struct node));
         if (new_node == NULL) {
            printf("Out of memory");
         }
         new_node->val = value;
         new_node->next = tmp->next;
         tmp->next = new_node;
         return;
      }
      tmp = tmp->next;
   }
}
void insert_before(struct node **head, int value, int before){
   struct node * new_node = NULL;
   struct node * tmp = *head;
   new_node = (struct node *)malloc(sizeof(struct node));
   if (new_node == NULL){
      printf("Out of memory");
      return;
   }
   new_node->val = value;
   if((*head)->val == before){
      new_node->next = *head;
      *head = new_node;
      return;
   }
   while(tmp && tmp->next) {
      if(tmp->next->val == before) {
         new_node->next = tmp->next;
         tmp->next = new_node;
         return;
      }
      tmp = tmp->next;
   }
   /*Before node not found*/
   free(new_node);
}
void main(){
   int count = 0, i, val, after, before;
   struct node * head = NULL;
   printf("Enter no: of elements: ");
   scanf("%d", &count);
   for (i = 0; i < count; i++){
      printf("Enter %dth element: ", i);
      scanf("%d", &val);
      insert_front(&head, val);
   }
   printf("starting list: ");
   print_list(head);
   printf("enter front element: ");
   scanf("%d", &val);
   insert_front(&head, val);
   printf("items after insertion: ");
   print_list(head);
   printf("enter last element: ");
   scanf("%d", &val);
   insert_end(&head, val);
   printf("items after insertion: ");
   print_list(head);
   printf("Enter an ele to insert in the list: ");
   scanf("%d", &val);
   printf("Insert after: ");
   scanf("%d", &after);
   insert_after(head, val, after);
   printf("List after insertion: ");
   print_list(head);
   printf("Enter an ele to insert in the list: ");
   scanf("%d", &val);
   printf("Insert before: ");
   scanf("%d", &before);
   insert_before(&head, val, before);
   printf("List after insertion: ");
   print_list(head);
}

Output

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

Enter no: of elements: 4
Enter 0th element: 1
Enter 1th element: 2
Enter 2th element: 3
Enter 3th element: 4
starting list: H->4->3->2->1->......
enter front element: 5
items after insertion: H->5->4->3->2->1->......
enter last element: 0
items after insertion: H->5->4->3->2->1->0->......
Enter an ele to insert in the list: 6
Insert after: 0
List after insertion: H->5->4->3->2->1->0->6->......
Enter an ele to insert in the list: 7
Insert before: 5
List after insertion: H->7->5->4->3->2->1->0->6->......

Bhanu Priya

Updated on: 24-Mar-2021

3K+ Views

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