java - 如何在java中生成大量素数？

Question

为了解决一个问题，我必须生成一个从 1 到 3000000 的素数列表，所以我尝试了几种方法来做到这一点，不幸的是都失败了......

第一次尝试:因为所有大于2的质数都是奇数，所以我先生成一个以3开头的奇数列表，叫做allOddNums。然后我生成一个名为 allComposite 的所有合数列表。然后我从 allOddNums 中删除 allComposite 中的所有数字以获得质数。这是我的代码:

/** Prime Numbers Generation
  * Tony
  */
import java.util.*;

public class PrimeNumG {
  public static void main(String[] args) {

    List <Long> allOddNums = new ArrayList<Long>();
    for (long i = 3; i < 200; i += 2) {
      allOddNums.add(i);
    }

    // composite number generator:
    List <Long> allComposite = new ArrayList<Long>();
    for (long a = 2; a < Math.round(Math.sqrt(3000000)); a += 2) {
      for (long b = 2; b < Math.round(Math.sqrt(3000000)); b += 2) {
        allComposite.add(a*b);
      }
    }

    // remove duplicated:
    Set <Long> hs = new HashSet<Long>();
    hs.addAll(allComposite);
    allComposite.clear();
    allComposite.addAll(hs);

    // remove all composite from allRealNums = allPrime
    allOddNums.removeAll(allComposite);
    allOddNums.add(0, (long)2);

    System.out.printf("%s ", allOddNums);
    Scanner sc = new Scanner(System.in);
    int times = sc.nextInt();

    for (int i = 0; i < times; i++) {
      int index = sc.nextInt();
      System.out.print(allOddNums.get(index) + " ");

    }
  }
}

在这种情况下，当我需要生成一些质数时，它工作正常。但是，如果我想生成直到 3000000，它就会失败(耗尽内存)。

第二次尝试:我在网上搜索并找到了一种名为埃拉托色尼筛法的算法。然后我首先生成 2, 3, 5, 7, 9...(所有奇数 + 2)，然后我删除 3 之后的每个第 3 个数字和 5 之后的每个第 5 个数字。代码如下:

/** Prime Number Generator
  * Tony
  */   
import java.util.*;

public class Solution61 {
  public static void main(String[] args) {
    List<Long> l1 = new ArrayList<Long> ();

    // l1 generator:   3 5 7 9 11 ... 
    for (long d = 3; d < 100; d += 2) {
      l1.add(d);
    }

    l1.add(1, (long)2); // 2 3 5 ...

    removeThird(l1); // rm 3rd after 3

    removeFifth(l1); // rm 5th after 5, now the l1 will be prime number

    Scanner sc = new Scanner(System.in);
    int times = sc.nextInt();
    for (int i = 0; i < times; i++) {
      int index = sc.nextInt();
      System.out.print(l1.get(index) + " ");


    }
  }


  /** removeThird : remove every 3rd number after 3
    * param List | return void
    */
  private static void removeThird(List<Long> l) {

    int i = 1;
    int count = 0;
    while (true) {


      if (count == 3) {
        l.remove(i);
        count = 1;

      }
      i ++;
      count ++;
      if (i > l.size()) {
        break;
      }
    }
  }

  /** removeThird : remove every 5th number after 5
    * param List | return void
    */
  private static void removeFifth(List<Long> l) {

    int i = 2;
    int count = 0;
    while (true) {


      if (count == 5) {
        l.remove(i);
        count = 1;
      }
      i ++;
      count ++;
      if (i > l.size()) {
        break;
      }
    }
  }

}

这仍然不能完成任务，因为它也耗尽了内存。

第三次尝试:我试着生成从 1 到 3000000，然后去掉每一个数是质数和另一个数的乘积。代码如下:

/** print all the prime numbers less than N
  * Tony
  */

public class primeGenerator {
  public static void main(String[] args) {
    int n = 3000000;
    boolean[] isPrime = new boolean[n];
    isPrime[0] = false; // because 1 is not a prime number

    for (int i = 1; i < n; i++) {
      isPrime[i] = true;
    } // we set 2,3,4,5,6...to true

    // the real number is always (the index of boolean + 1)

    for (int i = 2; i <= n; i++) {
      if (isPrime[i-1]) {
        System.out.println(i);
        for (int j = i * i; j < n; j += i /* because j is determined by i, so the third parameter doesn't mater*/) {
          isPrime[j-1] = false;
        }
      }
    }
  }
}

它仍然让我失望，我猜 3000000 真的是一个很大的数字吧？是否有任何简单而出色的新手友好方法来生成 3000000 以下的素数？谢谢!

第四次尝试:@jsheeran 下面的代码是您的答案的意思吗？当我达到 1093 时，它变得越来越慢，我的 IDE 仍然崩溃。如果我误解了你的方法，请告诉我，谢谢!

/** new approach to find prime numbers
  * Tony
  */
import java.util.*;

public class PrimeG {

  /** isPrime
    * To determine whether a number is prime by dividing the candidate number by each prime in that list
    */
  static List<Long> primes = new ArrayList<Long> ();

  private static void isPrime(long n) {
    boolean condition = true;
    for (int i = 0; i < primes.size(); i++) {
      if (n % primes.get(i) == 0) {
        condition = condition && false;
      }
    }
    if (condition) {
      findNextPrime(n);
    }
  }

  /** findNextPrime
    * expand the list of prime numbers 
    */
  private static void findNextPrime(long n) {
    primes.add(n);
  }





  public static void main(String[] args) {
    primes.add((long)2);
    primes.add((long)3);
    primes.add((long)5);
    primes.add((long)7);

    for (int i = 8; i < 3000000; i++) {
      isPrime(i);
      System.out.printf("%s", primes);
    }


  }
}

Answer 1

修复了埃拉托色尼筛法的实现(您的第三次尝试)。我相信它应该能满足您的需求。

public static void main (String[] args) throws java.lang.Exception {
    int n = 3000000;

    boolean[] isPrime = new boolean[n+1];
    for (int i = 2; i <= n; i++) {
        isPrime[i] = true;
    }

    for (int factor = 2; factor*factor <= n; factor++) {
        if (isPrime[factor]) {
            for (int j = factor; factor*j <= n; j++) {
                isPrime[factor*j] = false;
            }
        }
    }

    for (int i = 2; i <= n; i++) {
        if (isPrime[i]) System.out.println(i);
    }
}