Java 8如何使用Stream检查一个数是否被一个数除以

Question

How can i check if one range from numbers is divided by more than one number. For example if it is only 1 divisor this will work , but i don't know what to do when i have an array from numbers :

int limitNumber = 10;
int onlyOneDivisor = 2;
String input = "2 3";
int[] nums = Arrays.stream(input.split(" ")).mapToInt(Integer::parseInt).toArray();

//This works fine, but it's divided only by 2

IntStream.rangeClosed(0, limitNumber).filter(n -> (n % onlyOneDivisor == 0))
.forEach(x -> System.out.print(x + " "));

Answer 1

You can use filter , filtering by another stream over all the divisors, and using allMatch or anyMatch , depending on the exact use case:

int max = 10;
int[] divisors = {2, 3, 5};

// no divisor
IntStream.rangeClosed(0, max)
        .filter(n -> IntStream.of(divisors).allMatch(d -> n % d != 0))
        .forEach(System.out::println);

// any divisor
IntStream.rangeClosed(0, max)
        .filter(n -> IntStream.of(divisors).anyMatch(d -> n % d == 0))
        .forEach(System.out::println);

If you want to check whether a number has exactly two divisors (given a list of more than two), you could use an inner filter and combine that with count :

// exactly two divisors
IntStream.rangeClosed(0, max)
        .filter(n -> IntStream.of(divisors).filter(d -> n % d == 0).count() == 2)
        .forEach(System.out::println);

Note, however, that if you want to find prime numbers, you should probably not use a fixed list of divisors, but determine the divisors from the number under test:

IntStream.rangeClosed(0, max)
        .filter(n -> IntStream.range(2, n).allMatch(d -> n % d != 0))
        .forEach(System.out::println);

(Using n as the upper bound here is not very efficient; sqrt(n) would be enough, and only the odd numbers and two, and many more optimization.)

Answer 2

You could apply one filter per divisor.

IntStream stream = IntStream.rangeClosed(0, limitNumber);

for (int divisor: divisors) {
    stream = stream.filter(n -> n % divisor == 0);
}

stream.forEach(x -> System.out.print(x + " "));

To do it in one go you need to check for divisibility by all factors at once. It's a bit much to write out inline, so I recommend a helper method isDivisibleByAll (implementation left as an exercise).

IntStream.rangeClosed(0, limitNumber)
    .filter(n -> isDivisibleByAll(n, divisors))
    .forEach(x -> System.out.print(x + " "));

A smarter approach might be to compute the least common multiple of the factors, then just check for divisibility by that number.

Answer 3

I would use allMatch within filter :

String input = "2 3";

int[] divisors = Arrays.stream(input.split(" "))
    .mapToInt(Integer::parseInt)
    .toArray();

int limitNumber = 10;

IntStream.rangeClosed(0, limitNumber) // maybe 1 and not 0?
    .filter(n -> Arrays.stream(divisors).allMatch(d -> n % d == 0))
    .forEach(System.out::println); // 0 6

Here I'm streaming over the divisors array within filter , to check that all divisors actually divide the current number. If this condition is successful, the number remains in the stream and is finally printed.

Answer 4

"Divided by more than 1 number" means it's not prime , so:

IntStream.rangeClosed(0, limitNumber)
  .filter(this::isNotPrime)
  .forEach(x -> System.out.print(x + " "));

Then define a method:

private boolean isNotPrime(int n) {
    int s = (int)Math.sqrt(n);
    for (int i = 2; i < s; i++)
        if (n % 2 == 0) return true;
    return false;
}