素数的时间复杂度分析_分析判断n是否为素数,n≥0 算法的时间复杂度

找出素数的四种方法:

1、时间复杂度：O(n²)

• divisor <= Math.sqrt(number)

public class PremeNumber {
   
    public static void main(String[] args) {
   
        final int NUMBER_OF_PRIMES = 50 ;
        final int NUMBER_OF_PRIMES_PRE_LINE = 10;
        int count = 0;
        int number = 2;
        System.out.println("The first 50 prime numbers are : ");

        while (count < NUMBER_OF_PRIMES){
   
            boolean isPrime = true;
            for (int divisor = 2; divisor <= Math.sqrt(number); divisor++){
   
                if (number % divisor == 0){
   
                    isPrime = false;
                    break;
                }
            }
            if (isPrime){
   
                count++;
                if (count % NUMBER_OF_PRIMES_PRE_LINE == 0){
   
                    System.out.println(number);
                }
                else System.out.print(number+" ");
            }
            number++;
        }
    }
}

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2、时间复杂度：O(n√n)

1. 主要区别：

int  squareRoot = 0;         
if (squareRoot * squareRoot < number) squareRoot++;
   for (int divisor = 2; divisor <= squareRoot; divisor++){
   
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2.完整代码：

public