Printing prime number from 1 to 100

前端 未结 7 1547
情书的邮戳
情书的邮戳 2021-02-11 10:12

This program is supposed to output the prime numbers between 1 and 100. Will anyone please explain me the flow the programme below? I am having difficulty in writing the progra

相关标签:
7条回答
  • 2021-02-11 10:38

    If you split the various parts out into their own methods with appropriate names it becomes a bit easier to understand:

    for (int n = 1; n < 100; n++)
        if (isPrime(n))
            System.out.println(n);
    
    private boolean isPrime(int n) {
        for (int f = 2; f < n; f++) {
            if (isFactor(f, n))
                return false;
        }
        return true;
    }
    
    private boolean isFactor(int factor, int number) {
        return number % factor == 0;
    }
    

    This is also an area where Java 8 streams can make things a bit clearer:

    List<Integer> primes = IntStream.range(1, 100)
        .filter(this::hasNoFactors)
        .collect(Collectors.toList());
    
    private boolean hasNoFactors(int number) {
        return IntStream.range(2, number)
            .noneMatch(f -> number % f == 0);
    }
    

    Also note that this is a horribly inefficient algorithm. You don't need to check every possible factor from 2 to n, just the primes. You can also take advantage of multi-processor machines:

    List<Integer> primes = new ArrayList<>();
    IntStream.range(2, 100)
        .filter(n -> primes.parallelStream().noneMatch(p -> n % p == 0))
        .forEach(primes::add);
    
    0 讨论(0)
  • 2021-02-11 10:38

    public static ArrayList prime(int limit){

        ArrayList<Integer> primes = new ArrayList<>();
        for(int p = 2; p <= limit; p++) {
            int count = 0;
            for(int i=2; i < p; i++) {
    
                if (p%i == 0) {
                    count++;
                }
            }
    
            if (count == 0) {
                primes.add(p);
            }   
        }
        return primes;
    }
    
    0 讨论(0)
  • 2021-02-11 10:43

    How would you find a prime number with plain vanilla solution? If number is prime. It will not be a multiple of any number other than itself. So assume number is x. This number will not be divisible by any number when starting from 2 till x-1. Why start from 2 and not 1 because every number is divisible by 1. The above code is trying to replicate the same behavior. To find all primes between 1 till 99 (as per the loop):

    1. From 2 till number from (outer loop - 1)
    2. Try dividing the number and check if it's divisible. (remainder should be zero).
    3. If true number is not prime. Else number is prime.
    0 讨论(0)
  • 2021-02-11 10:45

    The number which is only divisible by itself and 1 is known as prime number. Here is the simplest version of the code for finding prime numbers between 1 to 100.

    import java.io.*;
    import java.util.*;
    class solution
    {
        public static void main(String args[])
        {
            Scanner scan = new Scanner(System.in);
            int n = 100;
            /*
                A prime number is a whole number greater than 1, whose only two whole-number factors are 1 and itself.
            */
            for(int i=2;i<=n;i++) // 1.So we are starting with initialization i = 2
            {
                int flag = 1;
                for(int j=2;j<=i/2;j++)  // 2.Try dividing the number by half check whether it divisible
                {
                    if(i%j==0) // 3. If the number is divisible by other number ->Not a prime Number
                    {
                        flag = 0;
                        break;
                    }
    
                }
                if(flag==1) // 4. If the number is not divisible by any other numbers but only by itself and 1 -> prime no
                {
                    System.out.print(i+" ");
                }
            }
        }
    }
    /*
        Output:
        2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97
    */
    
    0 讨论(0)
  • 2021-02-11 10:46
        public static List<Integer> getPrimeNumbers(int from , int to) {
        List<Integer> list = new ArrayList<>();
        for (int i = from;i <= to; i++) {
            int count = 0;
            for (int num = i; num>=1;num--) {
                if(i%num == 0){
                    count++;
                }
            }
            if(count ==2) {
                list.add(i);
            }
        }
        return list;
    }
    
    0 讨论(0)
  • 2021-02-11 10:49
    public class primenum {
    
        public static void main(String[] args) {
            for(int i=2;i<100;i++){
                int count=1;
                for(int j=2;j<i;j++){
                    if(i%j ==0){
                        count=0;
                        break;
                    }
    
                }if(count==1){
                    System.out.println(i);
    
                }
            }
    
        }
    
    0 讨论(0)
提交回复
热议问题