Insert a new first node by pointing it at the old head and then moving the head pointer.

Algorithm

The replay labels nodes by value, such as node(20), and never exposes object identity or memory addresses. This C# DSA implementation uses the same small chain as the rest of the DSA track.

old head The previous first node becomes the second node.
constant-time insert Only the new node and head pointer change.

Visual walkthrough

Head insertion changes only two references: the new node points at the old head, then head moves to the new node.

Step 1 - Old first node

Before insertion, head points at node(20).

Original chain before inserting 10 at the head.headnode(20)node(30)null

Step 2 - New node links to old head

Set new.next to the old first node before moving head.

node(10) is allocated and points at the old head node(20).headnode(10)newnode(20)old headnode(30)null

Step 3 - Head moves to the new node

The final chain has 10 first: 10 -> 20 -> 30 -> null.

After insertion, head points at node(10).headnode(10)node(20)node(30)null

Basic Implementation

basic.cs
using System;
using System.Text;

class Node {
    public int Value;
    public Node? Next;
    public Node(int value, Node? next = null) {
        Value = value;
        Next = next;
    }
}

class Program {
    static string Render(Node? head) {
        var outText = new StringBuilder();
        var cursor = head;
        while (cursor != null) {
            if (outText.Length > 0) outText.Append(" -> ");
            outText.Append(cursor.Value);
            cursor = cursor.Next;
        }
        return outText.Append(" -> null").ToString();
    }
    static void Main() {
        Node head = new Node(20, new Node(30));
        Node newHead = new Node(10);
        newHead.Next = head;
        head = newHead;
        Console.WriteLine(Render(head));
    }
}

Complexity

  • Time: O(1)
  • Space: O(1)

Implementation notes

  • Keep the explicit node and pointer/reference operations; array shortcuts hide the linked-list state this lesson is meant to replay.
  • The final output prints the chain in a deterministic a -> b -> null form for cross-language comparison.