reverse a linked list(iterative approach)

//returns new head pointer
Node reverse(Node head)
{
//empty list
if(head == null) return;

//all other cases, including only 1 node in the list
Node current = head.next;
Node prev = head;
Node tmp = null;


head.next = null;

while(current != null)
{
  tmp = current.next;
  current.next = prev;
  prev = current;
  current = tmp;
}

return prev;  //new head
}