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

Algorithm

Basic Implementation

basic.lua
local function node(value, next)
    return { value = value, next = next }
end

local function render(head)
    local parts = {}
    local cursor = head
    while cursor ~= nil do
        parts[#parts + 1] = tostring(cursor.value)
        cursor = cursor.next
    end
    return table.concat(parts, " -> ") .. " -> null"
end

local head = node(20, node(30, nil))
local new_head = node(10, nil)
new_head.next = head
head = new_head
print(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.