Algorithm Gossip: Eratosthenes筛选求质数_综合

package main01;

import java.util.ArrayList;

public class CoreJava05 {

/**
* @param args

* Algorithm Gossip: Eratosthenes筛选求质数解题思路:

过滤2的倍数，3的倍数，5的倍数，7的倍数.........N

public static void main(String[] args) {

// TODO Auto-generated method stub

//测试方法代码所用，

       System.out.println(Eratosthenes(30).toString());
   }

   public static ArrayList<Integer> Eratosthenes(int number) {
       ArrayList<Integer> list = new ArrayList<Integer>();
       int count = 0;
       int i = 0;
       int j = 0;
       int array[] = new int[number + 1];
       // 初始化数组不考虑1
       for (i = 2; i <= array.length - 1; i++) {
           array[i] = 1;
       }
       for (i = 2; i <= array.length - 1; i++) {
           // 从2开始凡是为2得倍数，3的倍数，4的倍数。。。。。。则将该值设为0
           for (j = i * i; j <= array.length - 1; j += i) {
               array[j] = 0;
           }
       }
       //将数组中为1的即为质数的下标保存在list集合中
       for (i = 2; i <= array.length - 1; i++) {
           if (array[i] == 1) {
               list.add(i);
           }
       }
       return list;
   }

}