(LeetCode)Count Primes --- 统计素数（质数）

Description:

Count the number of prime numbers less than a non-negative number, n.

Credits:

Special thanks to @mithmatt for adding this problem and creating all test cases.

Hint:

1. Let‘s start with a isPrime function. To determine if a number is prime, we need to check if it is not divisible by any number less than n. The runtime complexity of isPrimefunction
   would be O(n) and hence counting the total prime numbers up to n would be O(n2). Could
   we do better

解题分析：

为了寻求最小的时间复杂度，这里使用埃拉斯特尼筛选法，

学习链接：

http://blog.csdn.net/u012965373/article/details/52248039

# -*- coding:utf-8 -*-
__author__ = 'jiuzhang'
class Solution(object):
    def countPrimes(self, n):
        isPrime = [True] * max(n, 2)
        isPrime[0], isPrime[1] = False, False
        x = 2
        while x * x < n:
            if isPrime[x]:
                p = x * x
                while p < n:
                    isPrime[p] = False
                    p += x
            x += 1
        return sum(isPrime)