How do you remove the first node of a singly-linked list?

Check that the list is not empty, keep a temporary reference t to the old head, move head to head.next, cut the old node loose with t.next ← null, and decrement size — $O(1)$.

* t keeps the old first node in hand while head moves on to the second. *

Algorithm removeFirst()
  if head = null then indicate an error: the list is empty
  t ← head              { keep the old first node }
  head ← head.getNext() { head now points to the second node }
  t.setNext(null)       { unhook the old node from the list }
  size ← size − 1

Each line has a reason:

• The empty check comes first because head.getNext() on a null head would crash.
• t holds on to the old first node, so it can be unhooked and (in a real method) its element returned.
• t.setNext(null) removes the removed node's link into the list. It is not needed for correctness, but it stops the old node from keeping the rest of the list reachable through a stray reference, which helps the garbage collector.

Like addFirst, this is a constant number of reference updates: $O(1)$.