Insert a Node at Front of a Linked List

Last Updated : 21 Oct, 2025

Given a head of the linked list, Insert a new node at the beginning/start/front of the linked list.

Example:

Input: x = 1

Output: 1 -> 2 -> 3 -> 4 -> 5
Explanation: We can see that 1 is inserted at the beginning of the linked list.

Input: x = 1

Output: 1 -> 2 -> 10
Explanation: We can see that 1 is inserted at the beginning of the linked list.

Approach:

To insert a new node at the front, we create a new node and point its next reference to the current head of the linked list. Then, we update the head to be this new node. This operation is efficient because it only requires adjusting a few pointers.

Algorithm:

Make the first node of Linked List linked to the new node
Remove the head from the original first node of Linked List
Make the new node as the Head of the Linked List.

Below is the implementation of the approach:

C++

#include <iostream>
using namespace std;

// Define a node in the linked list
class Node {
public:
    int data;
    Node* next;

    // Constructor to initialize the node
    Node(int x) {
        data = x;
        next = nullptr;
    }
};

// Function to insert a new node
// at the beginning of the list
Node* insertAtFront(Node* head, int x) {
    Node* newNode = new Node(x);
    newNode->next = head;
    return newNode;
}

void printList(Node* head) {
    Node* curr = head;

    while (curr != nullptr) {
        cout << curr->data;
        if (curr->next != nullptr) {
            cout << " -> ";
        }
        curr = curr->next;
    }
    cout << endl;
}

int main() {
    // Create the linked list 2->3->4->5
    Node* head = new Node(2);
    head->next = new Node(3);
    head->next->next = new Node(4);
    head->next->next->next = new Node(5);

    // Insert a new node at the front of the list
    int x = 1;
    head = insertAtFront(head, x);

    printList(head);

    return 0;
}

#include <stdio.h>
#include <stdlib.h>

// Define a node in the linked list
struct Node
{
    int data;
    struct Node *next;
};

// Function to create a new node
struct Node *createNode(int x)
{
    struct Node *newNode = (struct Node *)malloc(sizeof(struct Node));
    newNode->data = x;
    newNode->next = NULL;
    return newNode;
}

// Function to insert a new node at the beginning of the list
struct Node *insertAtFront(struct Node *head, int x)
{
    struct Node *newNode = createNode(x);
    newNode->next = head;
    return newNode;
}

// Function to print the linked list
void printList(struct Node *head)
{
    struct Node *curr = head;
    while (curr != NULL)
    {
        printf("%d", curr->data);
        if (curr->next != NULL)
            printf(" -> ");
        curr = curr->next;
    }
    printf("\n");
}

int main()
{
    // Create the linked list 2->3->4->5
    struct Node *head = createNode(2);
    head->next = createNode(3);
    head->next->next = createNode(4);
    head->next->next->next = createNode(5);

    // Insert a new node at the front
    int x = 1;
    head = insertAtFront(head, x);

    // Print the updated linked list
    printList(head);

    return 0;
}

Java

class Node {
    int data;
    Node next;

    // Constructor to initialize the node
    Node(int x) {
        data = x;
        next = null;
    }
}

class GfG {

    // Function to insert a new node at 
    // the beginning of the list
    static Node insertAtFront(Node head, int x) {
        Node newNode = new Node(x);
        newNode.next = head;
        return newNode;
    }

    // Function to print the contents
    // of the linked list
    static void printList(Node head) {
        Node curr = head;

        while (curr != null) {
            System.out.print(curr.data);
            if (curr.next != null) {
                System.out.print(" -> ");
            }
            curr = curr.next;
        }
        System.out.println();
    }

    public static void main(String[] args) {
        // Create the linked list 2->3->4->5
        Node head = new Node(2);
        head.next = new Node(3);
        head.next.next = new Node(4);
        head.next.next.next = new Node(5);

        // Insert a new node at the 
        // front of the list
        int x = 1;
        head = insertAtFront(head, x);

        printList(head);
    }
}

Python

# Define a node in the linked list
class Node:
    def __init__(self, x):
        self.data = x
        self.next = None

# Function to insert a new node
# at the beginning of the list
def insertAtFront(head, x):
    newNode = Node(x)
    newNode.next = head
    return newNode

# Function to print the contents
# of the linked list
def printList(head):
    curr = head
    while curr is not None:
        print(curr.data, end="")
        if curr.next is not None:
            print(" -> ", end="")
        curr = curr.next
    print()

if __name__ == "__main__":
    # Create the linked list 2->3->4->5
    head = Node(2)
    head.next = Node(3)
    head.next.next = Node(4)
    head.next.next.next = Node(5)

    # Insert a new node at
    # the front of the list
    x = 1
    head = insertAtFront(head, x)

    # Print the updated list
    printList(head)

using System;

class Node {
    public int data;
    public Node next;

    // Constructor to initialize the node
    public Node(int x) {
        data = x;
        next = null;
    }
}

class GfG {

    // Function to insert a new node
    // at the beginning of the list
    static Node insertAtFront(Node head, int x) {
        Node newNode = new Node(x);
        newNode.next = head;
        return newNode;
    }

    // Function to print the contents
    // of the linked list
    static void printList(Node head) {
        Node curr = head;

        while (curr != null) {
            Console.Write(curr.data);
            if (curr.next != null) {
                Console.Write(" -> ");
            }
            curr = curr.next;
        }
        Console.WriteLine();
    }

    static void Main(string[] args) {
        // Create the linked list 2->3->4->5
        Node head = new Node(2);
        head.next = new Node(3);
        head.next.next = new Node(4);
        head.next.next.next = new Node(5);

        // Insert a new node at the front of the list
        int x = 1;
        head = insertAtFront(head, x);

        // Print the updated list
        printList(head);
    }
}

JavaScript

// Define a node in the linked list
class Node {
    constructor(x) {
        this.data = x;
        this.next = null;
    }
}

// Function to insert a new node 
// at the beginning of the list
function insertAtFront(head, x) {
    let newNode = new Node(x);
    newNode.next = head;
    return newNode;
}

// Function to print the contents 
// of the linked list
function printList(head) {
    let curr = head;
    while (curr !== null) {
        process.stdout.write(curr.data.toString());
        if (curr.next !== null) {
            process.stdout.write(" -> ");
        }
        curr = curr.next;
    }
    console.log();
}

// Driver code
function main() {
    // Create the linked list 2->3->4->5
    let head = new Node(2);
    head.next = new Node(3);
    head.next.next = new Node(4);
    head.next.next.next = new Node(5);

    // Insert a new node at 
    // the front of the list
    let x = 1;
    head = insertAtFront(head, x);

    // Print the updated list
    printList(head);
}

main();

Output

1 -> 2 -> 3 -> 4 -> 5

Time Complexity: O(1), We have a pointer to the head and we can directly attach a node and update the head pointer. So, the Time complexity of inserting a node at the head position is O(1).
Auxiliary Space: O(1)

animeshdey

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