[LeetCode] Count Primes

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

解题思路

采用Eratosthenes筛选法，依次分别去掉2的倍数，3的倍数，5的倍数，……，最后剩下的即为素数。

实现代码

//Rumtime:83ms
class Solution {
public:
    int countPrimes(int n) {
        int count = 0;
        bool *b = new bool[n];
        b[2] = true; //2是偶数，但不能被筛掉，需要特殊考虑
        for (int i = 3; i < n; i++)
        {
            if (i & 1)
            {
                b[i] = true; //奇数
            }
            else
            {
                b[i] = false;
            }
        }

        for (int i = 2; i < n; i++)
        {
            if (b[i])
            {
                count++;
                for (int j = 2; j * i < n; j++)
                {
                    b[i * j] = false;
                }
            }
        }

        delete [] b;
        return count;
    }
};