leetcode count prime

https://leetcode.com/problems/count-primes/

understanding

使用sieve methods. 首先排除0和1，然后从2开始逐个做标记，令p = 2，然后从$p^2$开始scan，排除掉所有 p的倍数multiples. 然后p increase to the most recent one that not been marked. 这里直接在最外面的while循环里面加个if判断就行，不要在while里面多弄一个while

这里要注意都是从$p*p$开始，因为p*(p-1)的话，已经被p-1的p倍标记过了

mycode:

class Solution(object):    def countPrimes(self, n):        """        :type n: int        :rtype: int        """        if n == 0 or n == 1:            return 0                        p = 2        flag = [1]* n        flag[0] = flag[1] = 0        while p*p < n:            if flag[p] == 1:#直接在这里判断下一个p是什么                i = p                while i * p < n:                    flag[i * p] = 0                    i += 1            p += 1                return sum(flag)

自己重写code

class Solution(object):    def countPrimes(self, n):        """        :type n: int        :rtype: int        """        if n < 2: return 0        flag = [1]*n        flag[0], flag[1] = 0, 0        p = 2        while p*p < n:            if flag[p*p] == 1:                i = p                while p*i < n:                    flag[p*i] = 0                    i += 1            p += 1        return sum(flag)