显示从 1 到 N 的所有质数的Java程序

对于给定的数 N，目的是找出从 1 到 N 的所有质数。

例子：

Input: N = 11
Output: 2, 3, 5, 7, 11

Input: N = 7
Output: 2, 3, 5, 7 

方法一：

• 首先，考虑给定的数字 N 作为输入。
• 然后应用 for 循环以迭代从 1 到 N 的数字。
• 最后，检查每个数字是否是质数，如果是质数，则使用蛮力方法打印它。

Java

// Java program to find all the
// prime numbers from 1 to N
class gfg {
 
    // Function to print all the
    // prime numbers till N
    static void prime_N(int N)
    {
        // Declaring the variables
        int x, y, flg;
 
        // Printing display message
        System.out.println(
            "All the Prime numbers within 1 and " + N
            + " are:");
 
        // Using for loop for traversing all
        // the numbers from 1 to N
        for (x = 1; x <= N; x++) {
 
            // Omit 0 and 1 as they are
            // neither prime nor composite
            if (x == 1 || x == 0)
                continue;
 
            // Using flag variable to check
            // if x is prime or not
            flg = 1;
 
            for (y = 2; y <= x / 2; ++y) {
                if (x % y == 0) {
                    flg = 0;
                    break;
                }
            }
 
            // If flag is 1 then x is prime but
            // if flag is 0 then x is not prime
            if (flg == 1)
                System.out.print(x + " ");
        }
    }
 
    // The Driver code
    public static void main(String[] args)
    {
        int N = 45;
 
        prime_N(N);
    }
}

输出

All the Prime numbers within 1 and 45 are:
2 3 5 7 11 13 17 19 23 29 31 37 41 43 

时间复杂度： O(N ² )

方法二：

• 首先，考虑给定的数字 N 作为输入。
• 然后应用 for 循环以迭代从 1 到 N 的数字。
• 最后，检查每个数字是否是素数，如果是素数，则使用平方根方法打印出来。

Java

// Java program to find all the
// prime numbers from 1 to N
class gfg {
 
    // Function to print all the
    // prime numbers till N
    static void prime_N(int N)
    {
        // Declaring the variables
        int x, y, flg;
 
        // Printing display message
        System.out.println(
            "All the Prime numbers within 1 and " + N
            + " are:");
 
        // Using for loop for traversing all
        // the numbers from 1 to N
        for (x = 2; x <= N; x++) {
 
            // Using flag variable to check
            // if x is prime or not
            flg = 1;
 
            for (y = 2; y * y <= x; y++) {
                if (x % y == 0) {
                    flg = 0;
                    break;
                }
            }
 
            // If flag is 1 then x is prime but
            // if flag is 0 then x is not prime
            if (flg == 1)
                System.out.print(x + " ");
        }
    }
 
    // The Driver code
    public static void main(String[] args)
    {
        int N = 45;
 
        prime_N(N);
    }
}

输出

All the Prime numbers within 1 and 45 are:
2 3 5 7 11 13 17 19 23 29 31 37 41 43 

时间复杂度： O(N ^3/2 )

方法三：

• 首先，考虑给定的数字 N 作为输入。
• 使用埃拉托色尼筛。

Java

// Java program to print all
// primes smaller than or equal to
// n using Sieve of Eratosthenes
 
class SieveOfEratosthenes {
    void sieveOfEratosthenes(int n)
    {
        // Create a boolean array
        // "prime[0..n]" and
        // initialize all entries
        // it as true. A value in
        // prime[i] will finally be
        // false if i is Not a
        // prime, else true.
        boolean prime[] = new boolean[n + 1];
        for (int i = 0; i <= n; i++)
            prime[i] = true;
 
        for (int p = 2; p * p <= n; p++) {
            // If prime[p] is not changed, then it is a
            // prime
            if (prime[p] == true) {
                // Update all multiples of p
                for (int i = p * p; i <= n; i += p)
                    prime[i] = false;
            }
        }
 
        // Print all prime numbers
        for (int i = 2; i <= n; i++) {
            if (prime[i] == true)
                System.out.print(i + " ");
        }
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int N = 45;
        System.out.println(
            "All the Prime numbers within 1 and " + N
            + " are:");
        SieveOfEratosthenes g = new SieveOfEratosthenes();
        g.sieveOfEratosthenes(N);
    }
}

输出

All the Prime numbers within 1 and 45 are:
2 3 5 7 11 13 17 19 23 29 31 37 41 43 

时间复杂度： O(n*log(log(n)))