如何使用 Java 中的一种方法将质数打印到 N

Question

I am trying to make a void method that prints prime numbers up to a given int argument. This is what I have and it's not working.

public class listPrimes {
    public static void main(String[] args) {
        printPrimes(1000);

    }

static void printPrimes(int max) {
    int counter = 0;
    for (int i = 2; i <= max; i++) {
        for (int n = 2; n < i; n++) {
            if (i % n == 0) {
                counter++;
            }
        }
        if (counter == 0) {
            System.out.println(i);
            counter = 0;
        }
    }
}
}

I was able to create the desired effect using 2 methods below but I want to do it with one. What is wrong with my code above?

public class listPrimes {
    public static void main(String[] args) {
        printPrimes(1000);
    }

private static void printPrimes(int max) {
    for (int i = 2; i <= max; i++) {
        if (primeCheck(i)) {
            System.out.println(i);
        }
    }
}

static boolean isPrime(int check) {
    for (int i = 2; i < check/2; i++) {
        if (check % i == 0) {
            return false;
        }
    }
    return true;
}
}

Answer 1

You never reset the value of counter in the outer loop . So, once it is increased, it will never be 0. So, just reset counter at the beginning of the outer loop: -

for (int i = 2; i <= max; i++) {
    counter = 0;
    for (int n = 2; n < i; n++) {
        if (i % n == 0) {
            counter++;
        }
    }
    if (counter == 0) {
        System.out.println(i);
    }
}

However, I would prefer to have the 2nd way, since it has divided the task in different method. So, both the methods have their defined roles. And you can use them at some other place easily. Just remember, the more you divide task amongst different methods, the more re-usability you get. It's always better to have one method do just one task.

Also, I would suggest to rename the method - primeCheck to isPrime , since it is returning a boolean value. So, just to follow the naming convention, isPrime will be a good name.

Answer 2

One variation, for better performance:

public static void main(String[] args) {
    printPrimes(100);
}

public static void printPrimes(int range) {
    for (int iCounter=1; iCounter<=range; ++iCounter) {
        if (iCounter <= 1) continue;
        if (iCounter == 2 || iCounter == 3) {
            System.out.println(iCounter);
            continue;
        }
        if (iCounter%2 == 0) continue;
        int iCounterSqrt = (int) Math.sqrt(iCounter);

        boolean bPrime = true;
        for (int iDenom=3; iDenom <= iCounterSqrt; iDenom += 2) {
            if (iCounter % iDenom == 0)
                bPrime = false;
        }

        if (bPrime) {
            System.out.println(iCounter);
        }
    }
}

Answer 3

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */

package javaapplication4;

/**
 *
 * @author sahadev & vignesh
 */
 class primecollection {


    public static void main(String[] args) {
     int r,s=0,k=0;
      int i,j;
      int a[] = new int[50];
      for(i=2;k<10;i++)
      {
        for(j=1;j<=i;j++)
        {
         r = i%j;

         if(r==0)
         {
          s++;
         }
         if(s==2)
         {
           a[k] = i; 
           k++;
         }
        }
        System.out.println(a[k]);
    }

}
}

Answer 4

If you are testing this with large numbers, using SQRT would be more efficient. http://www.wikihow.com/Check-if-a-Number-Is-Prime

    Scanner reader = new Scanner(System.in);
    System.out.println("Enter the a number");
    int num = reader.nextInt();
    int counter = 0;
    int root = 0;
    boolean prime_flag;

    if (2 <= num) {
        // 2 is the only even prime number
        counter++;
    }

    for (int i = 3; i < (num + 1); i++) {

        // test only for odd number
        if (i % 2 != 0) {
            prime_flag = true;
            root = (int) (Math.sqrt(i) + 1);

            for (int j = 3; j < (root + 1); j++) {
                if ((i % j == 0) && (i != j)) {

                    prime_flag = false;
                    break;
                }
            }

            if (prime_flag) {
                counter++;
            }

        }

    }

    System.out.println("Count of prime numbers upto " + num + " is "
            + counter);

Answer 5

Here is the code for getting prime numbers till 'n' in Javascript. It is a optimized code and you can refactor your code accordingly. Basic logic: For every number i check if it is divisible by any prime number we have already found till it is less than or equal to i/2.

function getPrimesTill(n){
  var i, j, len, limit, result = [];
  for(i=2;i<n;i++){
      limit = i/2;
      len = result.length;
      isPrime = true;
      for(j=0;len && result[j]<=limit;j++){
          if(i%result[j] == 0){
              isPrime = false;
              break;
          }
      }
      if(isPrime) result.push(i);
  }
  return result;
}
getPrimesTill(100);

Answer 6

You can reduce the time it takes the computer to find the prime numbers by following these steps!

• You only have to check if the new number is divisible by before primes
• You only have to do this up to the square root of the examined number
• You only have to check a number if it is a just number

So I'd do this:

public static void searchPrimes() {

    List<Long> primes = new ArrayList<Long>();

    System.out.println(2);

    loop:
    for(long x = 3; x < 100; x += 2) {

        int y = 0;

        while(primes.get(y) < Math.sqrt(x)) {

            if(x % primes.get(y) == 0) {

                continue loop;
            }

            y++;
        }

        primes.add(x);
        System.out.println(x);
    }
}

That's the cleanest and fastest way I found!

Answer 7

The simpliest way to n.

private static void nPrimeNumber(int n) {
    boolean result = true;
    for (int i = 2; i <= n; i++) {
        result = true;
        for (int j = 2; j < i; j++) {
            if (i % j == 0 ) {
                result = false;
                break;
            }
        }
        if(result){
            System.out.print(i+" ");
        }
    }
}

Answer 8

public class listPrimes {
public static void main(String[] args) {
    printPrimes(1000);

}

static void printPrimes(int max) 
{

    for(int k=2;k<=max;k++)
    { 
        int counter=0;
        for(int i=1;i<=k;i++)
        {
            if((k%i)==0)
            {
                counter++;
            }
        }
        if(counter==2)
        {
            System.out.println(k+"is prime");
        }
    }
}

}