C Program For Inserting A Node In A Linked List

We have introduced Linked Lists in the previous post. We also created a simple linked list with 3 nodes and discussed linked list traversal.
All programs discussed in this post consider the following representations of the linked list. 

C

// A linked list node
struct Node
{
  int data;
  struct Node *next;
};

In this post, methods to insert a new node in linked list are discussed. A node can be added in three ways 
1) At the front of the linked list 
2) After a given node. 
3) At the end of the linked list.

Add a node at the front: (4 steps process)
The new node is always added before the head of the given Linked List. And newly added node becomes the new head of the Linked List. For example, if the given Linked List is 10->15->20->25 and we add an item 5 at the front, then the Linked List becomes 5->10->15->20->25. Let us call the function that adds at the front of the list is push(). The push() must receive a pointer to the head pointer, because push must change the head pointer to point to the new node (See this)

Following are the 4 steps to add a node at the front.

C

// Given a reference (pointer to pointer) to
// the head of a list and an int,  inserts a
// new node on the front of the list.
void push(struct Node** head_ref,
          int new_data)
{
    // 1. Allocate node
    struct Node* new_node =
           (struct Node*) malloc(sizeof(struct Node));

    // 2. put in the data
    new_node->data  = new_data;

    // 3. Make next of new node as head
    new_node->next = (*head_ref);

    // 4. move the head to point to
    // the new node
    (*head_ref)    = new_node;
}

Time complexity of push() is O(1) as it does a constant amount of work.
Add a node after a given node: (5 steps process)
We are given a pointer to a node, and the new node is inserted after the given node.

C

// Given a node prev_node, insert a
// new node after the given prev_node
void insertAfter(struct Node* prev_node,
                 int new_data)
{
    // 1. Check if the given prev_node
    // is NULL
    if (prev_node == NULL)
    {
    printf("the given previous node cannot be NULL");
    return;
    }

    // 2. Allocate new node
    struct Node* new_node =
           (struct Node*) malloc(sizeof(struct Node));

    // 3. Put in the data
    new_node->data = new_data;

    // 4. Make next of new node as next
    // of prev_node
    new_node->next = prev_node->next;

    // 5. Move the next of prev_node
    // as new_node
    prev_node->next = new_node;
}

Time complexity of insertAfter() is O(1) as it does a constant amount of work.

Add a node at the end: (6 steps process)
The new node is always added after the last node of the given Linked List. For example if the given Linked List is 5->10->15->20->25 and we add an item 30 at the end, then the Linked List becomes 5->10->15->20->25->30. 
Since a Linked List is typically represented by the head of it, we have to traverse the list till the end and then change the next to last node to a new node.

Following are the 6 steps to add node at the end.

C

// Given a reference (pointer to pointer) to
// the head of a list and an int, appends a
// new node at the end
void append(struct Node** head_ref,
            int new_data)
{
    // 1. Allocate node
    struct Node* new_node =
           (struct Node*) malloc(sizeof(struct Node));

    // Used in step 5
    struct Node *last = *head_ref;

    // 2. Put in the data
    new_node->data  = new_data;

    // 3. This new node is going to be the
    //    last node, so make next of it as NULL
    new_node->next = NULL;

    // 4. If the Linked List is empty, then make
    // the new node as head
    if (*head_ref == NULL)
    {
       *head_ref = new_node;
       return;
    }

    // 5. Else traverse till the last node
    while (last->next != NULL)
        last = last->next;

    // 6. Change the next of last node
    last->next = new_node;
    return;
}

Time complexity of append is O(n) where n is the number of nodes in the linked list. Since there is a loop from head to end, the function does O(n) work. 
This method can also be optimized to work in O(1) by keeping an extra pointer to the tail of the linked list.

Following is a complete program that uses all of the above methods to create a linked list.

C

// A complete working C program to demonstrate
// all insertion methods on Linked List
#include <stdio.h>
#include <stdlib.h>

// A linked list node
struct Node
{
  int data;
  struct Node *next;
};

// Given a reference (pointer to pointer) to
// the head of a list and an int, inserts a
// new node on the front of the list.
void push(struct Node** head_ref,
          int new_data)
{
    // 1. Allocate node
    struct Node* new_node =
           (struct Node*) malloc(sizeof(struct Node));

    // 2. Put in the data
    new_node->data  = new_data;

    // 3. Make next of new node as head
    new_node->next = (*head_ref);

    // 4. move the head to point to
    //    the new node
    (*head_ref)    = new_node;
}

// Given a node prev_node, insert a
// new node after the given prev_node
void insertAfter(struct Node* prev_node,
                 int new_data)
{
    // 1. Check if the given prev_node
    //    is NULL
    if (prev_node == NULL)
    {
      printf("the given previous node cannot be NULL");
      return;
    }

    // 2. Allocate new node
    struct Node* new_node =
           (struct Node*) malloc(sizeof(struct Node));

    // 3. Put in the data
    new_node->data  = new_data;

    // 4. Make next of new node as next
    //    of prev_node
    new_node->next = prev_node->next;

    // 5. Move the next of prev_node
    //    as new_node
    prev_node->next = new_node;
}

// Given a reference (pointer to pointer) to
// the head of a list and an int, appends a
// new node at the end
void append(struct Node** head_ref,
            int new_data)
{
    // 1. Allocate node
    struct Node* new_node =
           (struct Node*) malloc(sizeof(struct Node));

    // Used in step 5
    struct Node *last = *head_ref;

    // 2. Put in the data
    new_node->data  = new_data;

    // 3. This new node is going to be the
    //    last node, so make next of it as NULL
    new_node->next = NULL;

    // 4. If the Linked List is empty, then make
    //    the new node as head
    if (*head_ref == NULL)
    {
       *head_ref = new_node;
       return;
    }

    // 5. Else traverse till the last node
    while (last->next != NULL)
        last = last->next;

    // 6. Change the next of last node
    last->next = new_node;
    return;
}

// This function prints contents of the
// linked list starting from head
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;

  // Insert 6.  So linked list
  // becomes 6->NULL
  append(&head, 6);

  // Insert 7 at the beginning.
  // So linked list becomes 7->6->NULL
  push(&head, 7);

  // Insert 1 at the beginning. So
  // linked list becomes 1->7->6->NULL
  push(&head, 1);

  // Insert 4 at the end. So linked list
  // becomes 1->7->6->4->NULL
  append(&head, 4);

  // Insert 8, after 7. So linked list
  // becomes 1->7->8->6->4->NULL
  insertAfter(head->next, 8);

  printf("Created Linked list is: ");
  printList(head);

  return 0;
}

Output:

 Created Linked list is:  1  7  8  6  4

Time Complexity: O(n) 
Auxiliary Space: O(1)

Please refer complete article on Linked List | Set 2 (Inserting a node) for more details!