JAVA：一种方法，该方法以int“ n”作为输入，并返回一个包含前n个素数的数组

Question

Here is what I got so far

public static int[] firstPrimeNumbers(int n) { 

    int[] k = new int[n];
    int m = 0;
    for (int i = 0; i < n; i++) { 

        if (isPrime(i)) { 
            k[m] = i;
            m++;
        }

    }
    return k;
}

Problem is, when I try to print the resulting array, I get a bunch of 0's at the end. For example when I took n = 10 , the program printed

2, 3, 5, 7, 0, 0, 0, 0, 0, 0

Hows that even possible? What am I doing wrong?

Answer 1

What happens here is fairly easy. k has size 10 and initially filled with zeros.

Then you check the first 10 natural numbers and check whether they are prime. So for each non-prime number you should get a 0

Maybe replace i<n by m<n but that depends a bit on what you want to achieve.

Answer 2

You are just counting until i == n (so only looking at number up to n ) - you need to keep incrementing i until m == n so you are considering whether numbers > n are prime.

If you used better variable names like nextPrimeIndex or primesFound instead of m this would probably be easier to spot.

Answer 3

You should modify your for loop. What you are doing is actually returning in an array of n elements the primitive numbers less than n

 public static int[] firstPrimeNumbers(int n) { 

      int[] k = new int[n];
      int nr = 2;
      int m = 0; 
      while(m<n){
         if(isPrime(nr)){ 
           k[m] = nr;
           m++;      
        }
         nr++;
      } 
     return k;
   }