イテレータの数を取得する「計算的に」迅速な方法はありますか?
int i = 0;
for ( ; some_iterator.hasNext() ; ++i ) some_iterator.next();
... CPUサイクルの無駄のようです。
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