[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:83msclass 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;    }};