¿Existe una forma rápida "computacionalmente" de obtener el recuento de un iterador?
int i = 0;
for ( ; some_iterator.hasNext() ; ++i ) some_iterator.next();
... parece una pérdida de ciclos de CPU.
to provide an implementation-independent method for access, in which the user does not need to know whether the underlying implementation is some form of array or of linked list, and allows the user go through the collection without explicit indexing.
penguin.ewu.edu/~trolfe/LinkedSort/Iterator.html