How to count the number of trailing zeroes in an integer using Java 8 Stream/Lambda?
Basically the logic should be: keep the integer dividing by 10 as long as the remainder is 0 (the quotient will be supplied to the next division) and count the occurrence(s).
e.g.
12300 % 10 == 0 true
1230 % 10 == 0 true
123 % 10 == 0 false
Answer: 2
Note: I prefer not to involve String here :-)
If this is a purely hypothetical question, here is a purely hypothetical answer of how you can do it:
static int countZeroes(int value) {
if(value == 0) // we need to handle this case explicitly
return 1;
IntStream s = IntStream.iterate(value, v -> v / 10);
return (int) takeWhile(s, v -> v > 0 && v % 10 == 0)
.count();
}
It uses a helper function takeWhile that is available in Java 9 but not in Java 8 so has to be emulated like this:
// In Java 9 there is a standard takeWhile
// https://docs.oracle.com/javase/9/docs/api/java/util/stream/Stream.html#takeWhile-java.util.function.Predicate-
// but in Java 8 I have to emulate it
static IntStream takeWhile(IntStream s, final IntPredicate pr) {
final Spliterator.OfInt origSp = s.spliterator();
Spliterator.OfInt filtered = new Spliterators.AbstractIntSpliterator(origSp.estimateSize(), 0) {
boolean lastPredicate = true;
@Override
public boolean tryAdvance(final IntConsumer action) {
if (!lastPredicate)
return false;
origSp.tryAdvance((int v) -> {
lastPredicate = pr.test(v);
if (lastPredicate) {
action.accept(v);
}
});
return lastPredicate;
}
};
return StreamSupport.intStream(filtered, false);
}
The idea is that
IntStream.iterate(value, v1 -> v1 / 10).takeWhile(v -> v > 0)
should generate a stream of cutting digits at the end one by one and then you can apply takeWhile(v -> v % 10 == 0).count() to count the number of zeros and finally you can merge those two takeWhiles into one.