【LEETCODE】204-Count Primes

Description:

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

思路：

The Sieve of Eratosthenes is one of the most efficient ways to find all prime numbers up ton. 

埃拉托斯特尼筛法 (http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes)

1. Create a list of consecutive integers from 2 through n: (2, 3, 4, ..., n).
2. Initially, let p equal 2, the smallest prime number.
3. Starting from p, enumerate its multiples by counting to n in increments of p, and mark them in the list (these will be 2p, 3p, 4p, ... ; the p itself should not be marked).
4. Find the first number greater than p in the list that is not marked. If there was no such number, stop. Otherwise, let p now equal this new number (which is the next prime), and repeat from step 3.

class Solution(object):    def countPrimes(self, n):        """        :type n: int        :rtype: int        """                isprime=[True]*max(n,2)             #1~n先全部设定为true                isprime[0]=isprime[1]=False         #1和2初始化为false                i=2                 while i*i<n:                        #因n＝i＊j，且i<=j，则 i＊i<span style="font-family: Arial, Helvetica, sans-serif;"><n</span>            if isprime[i]:                  #只需要check还不是false的i                p=i*i                       #比i小的部分i＊1～i＊(i－1)已经被check过了，所以直接判断i＊i的                while p<n:                    isprime[p]=False                    p+=i                    #p+i            i+=1                return sum(isprime)