java 8 中的质数
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Prime number in java 8
提问by user3310115
I was trying to write a simple prime number program in Java 8. Below is the program. I wanted to reduce the code in isPrime()as well. Is there something that filters the elements from 2to n/2, and then apply filter for n%i == 0which would make isPrimeirrelevant?
我试图用 Java 8 编写一个简单的素数程序。下面是程序。我也想减少代码isPrime()。是否有什么东西可以过滤元素从2to n/2，然后应用过滤器，n%i == 0这会变得isPrime无关紧要？
import static java.util.stream.Collectors.toList;
import java.util.Arrays;
import java.util.List;
import java.util.function.Predicate;
public class Stream1 {
public static void main(String[] args) {
List<Integer> numbers = Arrays.asList(1, 2, 3, 4, 5, 6, 7, 8, 9, 20);
// Prime number
System.out.println(numbers.stream()
.filter(Stream1::isPrime)
.collect(toList()));
}
public static boolean isPrime(int number) {
for (int i = 2; i <= number / 2; i++) {
if (number % i == 0) {
return false;
}
}
return true;
}
}
回答by Shashwat Kumar
IntStreamcan be used to generate integer stream
IntStream可用于生成整数流
public static boolean isPrime(int number) {
return !IntStream.rangeClosed(2, number/2).anyMatch(i -> number%i == 0);
}
or
或者
public static boolean isPrime(int number) {
return IntStream.rangeClosed(2, number/2).noneMatch(i -> number%i == 0);
}
回答by rossum
Your isPrime()is inefficient. First, you do not need to divide by any even numbers greater than 2, since an initial division by 2 will catch all even non-primes. Second, you terminate your loop at number / 2instead of the more efficient sqrt(number).
你isPrime()的效率低下。首先，您不需要除以任何大于 2 的偶数，因为初始除以 2 将捕获所有偶数非素数。其次，您将循环终止于number / 2而不是更有效的sqrt(number)。
You could rewrite your method something like this:
你可以像这样重写你的方法：
public static boolean isPrime(int number) {
// Even numbers
if (number % 2 == 0) {
return number == 2;
}
// Odd numbers
int limit = (int)(0.1 + Math.sqrt(number));
for (int i = 3; i <= limit; i += 2) {
if (number % i == 0) {
return false;
}
}
return true;
}
The Sieve of Eratosthenes would be more efficient still, but that might be overkill for a relatively small problem.
Eratosthenes 的筛子仍然会更有效，但对于一个相对较小的问题来说这可能是过度的。
回答by Eugene
As suggest by @rossum you can use the famous Sieve of Eratosthenesfor this and it will compute the primes pretty fast.
正如@rossum 所建议的，您可以为此使用著名的 Eratosthenes 筛，它会非常快地计算素数。
private static BitSet primes(int limit) {
BitSet bitSet = new BitSet(limit);
bitSet.set(0, false);
bitSet.set(1, false);
bitSet.set(2, limit, true);
for (int i = 2; i * i < limit; ++i) {
if (bitSet.get(i)) {
int j = i;
int x = 2;
while (j < limit) {
j = i * x;
bitSet.set(j, false);
++x;
}
}
}
return bitSet;
}
回答by Deep
You can achieve the desired using predicate as well.
您也可以使用谓词实现所需的效果。
import java.util.Arrays;
import java.util.List;
import java.util.function.IntPredicate;
import java.util.stream.Collectors;
import java.util.stream.IntStream;
public class PrimeUsingStream {
public static boolean isPrime(int i) {
IntPredicate isDivisible = index -> i % index == 0;
return i > 1 && IntStream.range(2, i).noneMatch(isDivisible);
}
public static void main(String[] args)
{
//System.out.println(printPrime(200));
List<Integer> numbers = Arrays.asList(1, 2, 3, 4, 5, 6, 7, 8, 9, 20,23);
// Prime number
System.out.println(numbers.stream()
.filter(PrimeUsingStream::isPrime)
.collect(Collectors.toList()));
}
}
回答by S. Zror
You can use stream as this test below:
您可以使用流作为下面的测试：
@Test
public void generatePrimeNumberListByStream(){
List<Integer> primeNumbers =
IntStream
.range(2,30)
.filter(number -> IntStream.range(2,number)
.noneMatch(divider -> number % divider == 0))
.boxed()
.collect(Collectors.toList());
assertThat(primeNumbers, contains(2,3,5,7,11,13, 17,19, 23, 29));
}
