Question

我试图弄清楚我将如何操作列表以查找所有由用户提供的数字的质数，我有一个列表步骤，我试图遵循这些步骤

创建并填充可能的素数列表

基本上是一个arrayList，其中包含直到提供的数字的所有数字，我已经完成了一部分

创建素数列表

我要把那部分放下来

虽然仍有可能的数字

也就是说，当可能的素数列表不为空时。

将可能性列表中的第一个数字添加到素数列表

也把那部分放下来

从可能的素数列表中删除它及其倍数

这是我开始发呆的地方，我以为我的那部分下降了，但是我收到一个错误，我不知道为什么。

打印素数

打印素数列表，本质上只是 System.out.println（primes）;

此刻我的代码

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Scanner;

public class Sieve {
public static void main(String[] args) {
    int maxNum;
    String howItsGoing, greatDetail;
    Scanner scnr = new Scanner(System.in);
    Scanner scnr2 = new Scanner(System.in);

    // get upper limit
    System.out.print("What's the biggest number I should check? ");
    maxNum = scnr.nextInt();

    // check for verbose mode
    System.out.print("Shall I tell you how it's going? ");
    howItsGoing = scnr2.nextLine().toUpperCase();

    System.out.print("Shall I tell you in great detail? ");
    greatDetail = scnr2.nextLine().toUpperCase();

    // create and fill list of possible primes
    List<Integer> nums = new LinkedList<>();

    for (int i = 2; i <= maxNum; i++) {
        nums.add(i);         
    }

    // create list for the primes
    List<Integer> primes = new ArrayList<>();
    // while there are still possible numbers

    // add the first number from the list of possibles to the list of primes
    for(int i=2; i<=maxNum; i++) {
        if(2 % i == 0) {
            nums.remove((Integer) i);
            primes.add((Integer) i);
        }
    }

        // remove it and its multiples from the list of possible primes

    // print the prime numbers
    System.out.println("Primes up to " + maxNum);
    System.out.println(nums);
    System.out.println(primes);


}

}

忽略howItsGoing字符串和greatDetail，我将在以后添加。

我将如何使该程序正常工作，其他所有问题都在布尔数组中有解决方案，这不是我想要的。有什么想法吗？

输出

What's the biggest number I should check? 9
Shall I tell you how it's going? n
Shall I tell you in great detail? n
Primes up to 9
[3, 4, 5, 6, 7, 8, 9]
[2]

Answer 1

我强调了这些错误：

    // remove it and its multiples from the list of possible primes
    for(int i=0; i<=maxNum; i++) {
        if(i % 2 == 0) {  // first bug
            nums.remove(i); // second bug
        }
    }

第一个错误是i％2 == 0检查i是否为2的倍数，但应检查i是否为当前素数的倍数。

第二个错误是

nums.remove(i);

是模棱两可的。 ArrayList<Integer>声明了两种不同的删除方法：remove(int)删除列表中的第i个条目，而remove(Integer)删除等于i的条目。由于int可以转换为Integer，因此这两种方法都可以匹配您的参数类型，但是remove(int)更适合参数类型，因此可以使用，尽管您可能想要remove（整数）...您可以通过强制转换参数来解决此问题：

nums.remove((Integer) i);

这应该使代码正常工作，但是您很快就会意识到该代码相当慢。这是因为remove(Integer)实际上是一个相当昂贵的操作，因为它涉及遍历整个List直到找到要删除的Integer为止。也就是说，对于您淘汰的每个主要候选人，您都会与所有其他主要候选人互动。由于其中有很多，因此您的代码将非常慢。

解决方案是选择一种数据结构，以更有效地支持按值删除。这就是为什么每个人在实现此算法时都使用boolean[]的原因。

Answer 2

弄清楚了，这就是代码的样子

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Scanner;

public class Sieve {
public static void main(String[] args) {
    int maxNum;
    boolean possible = true;
    String howItsGoing, greatDetail;
    Scanner scnr = new Scanner(System.in);
    Scanner scnr2 = new Scanner(System.in);

    // get upper limit
    System.out.print("What's the biggest number I should check? ");
    maxNum = scnr.nextInt();

    // check for verbose mode
    System.out.print("Shall I tell you how it's going? ");
    howItsGoing = scnr2.nextLine().toUpperCase();

    if (howItsGoing.startsWith("N")) {
        greatDetail = "N";
    } else {
        System.out.print("Shall I tell you in great detail? ");
        greatDetail = scnr2.nextLine().toUpperCase();
    }

    // create and fill list of possible primes
    List<Integer> nums = new LinkedList<>();

    for (int i = 2; i <= maxNum; i++) {
        nums.add(i);
    }

    // create list for the primes
    List<Integer> primes = new ArrayList<>();
    // while there are still possible numbers
    while (possible) {
        // add the first number from the list of possibles to the list of
        // primes

        primes.add(nums.get(0));

        if (howItsGoing.startsWith("Y")) {
            System.out.println();
            System.out.println();
            System.out.print("Found prime: ");
            System.out.printf("%1d ", nums.get(0));
            System.out.println();
        }
        // remove it and its multiples from the list of possible primes
        int divisor = nums.get(0);

        nums.remove(nums.get(0));

        for (int i = divisor; i <= maxNum; i++) {

            if (i % divisor == 0) {
                if (greatDetail.startsWith("Y")) {
                    System.out.println(
                            "    Removing " + i + " from possibles");
                }
                nums.remove((Integer) i);
            }

        }
        System.out.println();
        if (nums.size() > 0) {
            if (howItsGoing.startsWith("Y")) {
                System.out.print("Possibles:\n    ");
                for (int i = 0; i < nums.size(); i++) {
                    System.out.printf("%6d ", nums.get(i));
                }

            }
        }
        if (nums.size() < 1) {
            possible = false;
        }
    }

    // print the prime numbers
    System.out.println();
    System.out.println("Primes up to " + maxNum);
    for (int i = 0; i < primes.size(); i++) {
        System.out.printf("%6d ", primes.get(i));
    }

}
}

输出

What's the biggest number I should check? 20
Shall I tell you how it's going? yes
Shall I tell you in great detail? yes


Found prime: 2 
Removing 2 from possibles
Removing 4 from possibles
Removing 6 from possibles
Removing 8 from possibles
Removing 10 from possibles
Removing 12 from possibles
Removing 14 from possibles
Removing 16 from possibles
Removing 18 from possibles
Removing 20 from possibles

Possibles:
     3      5      7      9     11     13     15     17     19 

Found prime: 3 
Removing 3 from possibles
Removing 6 from possibles
Removing 9 from possibles
Removing 12 from possibles
Removing 15 from possibles
Removing 18 from possibles

Possibles:
     5      7     11     13     17     19 

Found prime: 5 
Removing 5 from possibles
Removing 10 from possibles
Removing 15 from possibles
Removing 20 from possibles

Possibles:
     7     11     13     17     19 

Found prime: 7 
Removing 7 from possibles
Removing 14 from possibles

Possibles:
    11     13     17     19 

Found prime: 11 
Removing 11 from possibles

Possibles:
    13     17     19 

Found prime: 13 
Removing 13 from possibles

Possibles:
    17     19 

Found prime: 17 
Removing 17 from possibles

Possibles:
    19 

Found prime: 19 
Removing 19 from possibles


Primes up to 20
 2      3      5      7     11     13     17     19

Eratosthenes筛（使用链接列表）

2 个答案: