Java implements three methods for finding primes less than n

Prime concept

Prime, also known as the prime, to point to in a is greater than 1 (link: http://zh.wikipedia.org/wiki/%E8%87%AA%E7%84%B6%E6%95%B0), in addition to 1 and the integer itself, cannot be other natural number (link: http://zh.wikipedia.org/wiki/%E6%95%B4%E9%99%A4) number (can also be defined as only 1 and itself two (link: http://zh.wikipedia.org/wiki/%E5%9B%A0%E6%95%B8).
The smallest prime number is 2, is also a prime number, the only (link: http://zh.wikipedia.org/wiki/%E5%81%B6%E6%95%B0); Other prime Numbers are (link: http://zh.wikipedia.org/wiki/%E5%A5%87%E6%95%B0). There are infinitely many primes, so there is no largest prime.

1. Solve according to the definition:
It is also the most stupid way, which is less efficient:

package test.ms;

public class FindPrime {
	 // find the prime  between 1 to 1000;
	public static void main(String[] args) {
		 printPrime(1000);
	}
	public static void  printPrime(int n){
		
		for(int i = 2; i < n ; i++){
			
			int count = 0;
			
			for(int j = 2 ; j<=i; j++){
				
				if(i%j==0){
					count++;
				}
				if(j==i & count == 1){
					System.out.print(i+" ");
				}
				if(count > 1){
					break;
				}
			}
			
			
		}
		
	}

}

2: square root:

package test.ms;

public class Prime { 
	
	public static void main(String[] args) {
		
		for(int j = 2; j<1000; j++){
			if(m(j)){
				System.out.print(j+" ");
			}
		}
	}
	
	public static boolean  m(int num){
	
		for(int j = 2; j<=Math.sqrt(num);j++){
			if(num%j == 0){
				return false;
			}
		}
		
		return true;
	}

}

3: find a pattern (from a forum discussion)

The smallest prime is 2, which is the only even number of primes; All the other primes are odd. There are infinitely many primes, so there is no largest prime.

package test.ms;

import java.util.ArrayList;
import java.util.List;

public class Primes {
		 
	  public static void main(String[] args) {
	  	
	    //For prime
	    List<Integer> primes = getPrimes(1000);
	 
	    //The output
	    for (int i = 0; i < primes.size(); i++) {
	      Integer prime = primes.get(i);
	      System.out.printf("%8d", prime);
	      if (i % 10 == 9) {
	        System.out.println();
	      }
	    }
	  }
	 
	  
	  private static List<Integer> getPrimes(int n) {
	    List<Integer> result = new ArrayList<Integer>();
	    result.add(2);
	 
	    for (int i = 3; i <= n; i += 2) {
	      if (!divisible(i, result)) {
	        result.add(i);
	      }
	    }
	 
	    return result;
	  }
	 
	  
	  private static boolean divisible(int n, List<Integer> primes) {
	    for (Integer prime : primes) {
	      if (n % prime == 0) {
	        return true;
	      }
	    }
	    return false;
	  }
	}

The first and second methods are both very simple:
The third method illustrates the property of a prime number: of all primes, only 2 is even.
A number is not prime if it is divisible by its previous prime.