Linked List | Set 2 (Inserting a node)

• Difficulty Level : Easy
• Last Updated : 28 Jun, 2021

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 linked list. 

• C++
• C
• Java
• Python
• C#
• Javascript

// A linked list node

class Node

{

    public:

    int data;

    Node *next;

};

// This code is contributed by rathbhupendra

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.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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++
• C
• Java
• Python
• C#
• Javascript

/* 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(Node** head_ref, int new_data)

{

    /* 1. allocate node */

    Node* new_node = new 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;

}

// This code is contributed by rathbhupendra

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++
• C
• Java
• Python
• C#
• Javascript

// Given a node prev_node, insert a

// new node after the given 

// prev_node

void insertAfter(Node* prev_node, int new_data) 

{

    // 1. Check if the given prev_node is NULL

    if (prev_node == NULL) 

    { 

        cout << "the given previous node cannot be NULL"; 

        return; 

    }

    // 2. Allocate new node

    Node* new_node = new 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; 

}

// This code is contributed by anmolgautam818

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++
• C
• Java
• Python
• C#
• Javascript

// Given a reference (pointer to pointer) to the head 

// of a list and an int, appends a new node at the end

void append(Node** head_ref, int new_data) 

{ 

    // 1. allocate node

    Node* new_node = new Node();

    // Used in step 5

    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 code is contributed by anmolgautam818

Time complexity of append is O(n) where n is the number of nodes in 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 linked list/

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

• C++
• C
• Java
• Python
• C#
• Javascript

// A complete working C++ program to demonstrate

//  all insertion methods on Linked List

#include <bits/stdc++.h>

using namespace std;

// A linked list node

class Node

{

    public:

    int data;

    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(Node** head_ref, int new_data)

{

    /* 1. allocate node */

    Node* new_node = new 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(Node* prev_node, int new_data)

{

    /*1. check if the given prev_node is NULL */

    if (prev_node == NULL)

    {

        cout<<"the given previous node cannot be NULL";

        return;

    }

    /* 2. allocate new node */

    Node* new_node = new 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(Node** head_ref, int new_data)

{

    /* 1. allocate node */

    Node* new_node = new Node();

    Node *last = *head_ref; /* used in step 5*/

    /* 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

// linked list starting from head

void printList(Node *node)

{

    while (node != NULL)

    {

        cout<<" "<<node->data;

        node = node->next;

    }

}

/* Driver code*/

int main()

{

    /* Start with the empty list */

    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);

    cout<<"Created Linked list is: ";

    printList(head);

    return 0;

}

// This code is contributed by rathbhupendra

Output:

Created Linked list is:  1  7  8  6  4