素数的时间复杂度分析

找出素数的四种方法:

复杂度	主要区别
1. 时间复杂度：O(n2)	divisor <= Math.sqrt(number)
2. 时间复杂度：O(n√n)	divisor <= squareRoot，其中squareRoot2
3. 时间复杂度：O(n√n / logn)	for (int k=0; k
4. 时间复杂度：O(n√n / logn)	for (int i=k; i<= n/k; i++){ //执行了 n/k-(k-1) primes[k * i] = false;

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++;
        }
    }
}

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++){

2.完整代码：

public class PrimeNumbers {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        System.out.println("Find all prime numbers <= n, enter n: ");
        int n = input.nextInt();
        final  int NUMBERS_PRE_LINE = 10;
        int count = 0;
        int number = 2;

        System.out.println("The prime numbers are: ");
        int squareRoot = 0;
        while (number <= n){
            boolean isPrime = true;
            //时间复杂度√n。
//            for (int divisor = 2; divisor <= (int) (Math.sqrt(number)); divisor++){
            //时间复杂度n√n。
            if (squareRoot * squareRoot < number) squareRoot++;
            for (int divisor = 2; divisor <= squareRoot; divisor++){

                if (number % divisor == 0){
                    isPrime = false;
                    break;
                }
            }

            if (isPrime){
                count++;
                if (count % NUMBERS_PRE_LINE == 0)
                    System.out.printf("%7d\n",number);
                else System.out.printf("%7d",number);
            }
            number++;
        }
        System.out.println("\n"+ count+" prime less than or equal to "+ n);

    }
}

3.时间复杂度分析：

3、时间复杂度：O(n√n / logn)

1. 主要区别：

//A list to hold prime numbers
java.util.ArrayList<Integer> list = new java.util.ArrayList<Integer>();
//循环体内
	for (int k=0; k<list.size() && list.get(k) <= squareRoot ;k++){...}

2. 完整代码：

public class EfficientPrimeNumbers {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        System.out.println("Find all prime numbers <= n, enter n: ");
        int n = input.nextInt();

        //A list to hold prime numbers
        java.util.ArrayList<Integer> list = new java.util.ArrayList<Integer>();

        final  int NUMBERS_PRE_LINE = 10;
        int count = 0;
        int number = 2;
        int squareRoot = 0;
        System.out.println("The prime numbers are: ");

        while (number <= n){
            boolean isPrime = true;
//            for (int divisor = 2; divisor <= (int) (Math.sqrt(number)); divisor++){

            if (squareRoot * squareRoot < number) squareRoot++;
            for (int k=0; k<list.size() && list.get(k) <= squareRoot ;k++){
                if (number%list.get(k) == 0){
                    isPrime = false;
                    break;
                }
            }
            if (isPrime){
                count++;
                list.add(number);
                if (count % NUMBERS_PRE_LINE == 0)
                    System.out.printf("%7d\n",number);
                else System.out.printf("%7d",number);
            }
            number++;
        }
        System.out.println("\n"+ count+" prime less than or equal to "+ n);
    }
}

3.时间复杂度分析：

4.算法介绍：

4、 时间复杂度：O(n√n / logn)

1.主要区别：

 //Initial primes[i] to true
 for (int i=0; i<primes.length; i++){
     primes[i] = true;
 }

 for (int k=2; k<= n/k; k++){
     if (primes[k]){
         for (int i=k; i<= n/k; i++){  //执行了 n/k-(k-1)次
             primes[k * i] = false;
         }
     }
 }

2.完整代码：

public class SieveOfEratosthenes {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        System.out.println("Find all prime numbers <= n, enter n: ");
        int n = input.nextInt();

        boolean[] primes = new boolean[n+1];

        //Initial primes[i] to true
        for (int i=0; i<primes.length; i++){
            primes[i] = true;
        }

        for (int k=2; k<= n/k; k++){
            if (primes[k]){
                for (int i=k; i<= n/k; i++){  //执行了 n/k-(k-1)次
                    primes[k * i] = false;
                }
            }
        }
        int count = 0;
        for (int i=2;i<primes.length;i++){
            if (primes[i]){
                count++;
                if (count % 10 == 0)
                    System.out.printf("%7d\n",i);
                else System.out.printf("%7d",i);
            }
        }
        System.out.println("\n"+ count+" prime less than or equal to "+ n);
    }
}

3.时间复杂度分析：

4.算法介绍：

5、常见的递归函数