题目链接:count-primes
Description:
Count the number of prime numbers less than a non-negative number, n
public class Solution {
	public int countPrimes(int n) {
		if(n <= 2) return 0;
		List<Integer> primes = new ArrayList<Integer>();
		primes.add(2);
		for (int i = 3; i < n; i += 2) {
			int sqrt_i = (int) Math.sqrt(i);
			for (int j = 0; i % primes.get(j) != 0; j++) {
				if (primes.get(j) > sqrt_i) {
					primes.add(i);
					break;
				}
			}
		}
		return primes.size();
    }
}原文:http://blog.csdn.net/ever223/article/details/45341267