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

Algorithm

Basic Implementation

basic.kt
class Node(val value: Int, var next: Node? = null)

fun render(head: Node?): String {
    val parts = mutableListOf<String>()
    var cursor = head
    while (cursor != null) {
        parts.add(cursor.value.toString())
        cursor = cursor.next
    }
    return parts.joinToString(" -> ") + " -> null"
}

fun main() {
    var head = Node(20, Node(30))
    val newHead = Node(10)
    newHead.next = head
    head = newHead
    println(render(head))
}

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

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