Description:

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

Link:

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

解題方法：

Sieve of Eratosthenes

Time Complexity：

O(N*logN)

完整代碼：

int countPrimes(int n) 
    {
        if(n < 3)
            return 0;
        vector<bool> prime(n, true);
        int cnt = 0;
        for(int i = 2; i < n; i++)
        {
            if(!prime[i])
                continue;
            int ftr = 2;
            while(i * ftr < n)
                prime[i * ftr++] = false;
            cnt++;
        }
        return cnt;
    }