/*
Omitted Enqueue and other set up
*/
public E dequeue(Stacks)
{
if(s.size()==1)
return s.pop()
E tmp = s.pop();
E rez = popElem(s);
s.push(tmp);
return rez;
}
Anonym
3. Jan. 2015
The Roman's answer is almost good, if you remove the second copyElements call from dequeue then you get amortized O(1) time for both operations.