[LeetCode] Count Primes，leetcodeprimes

[LeetCode] Count Primes，leetcodeprimes

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

解題思路

採用Eratosthenes篩選法，依次分別去掉2的倍數，3的倍數，5的倍數，……，最後剩下的即為素數。

實現代碼

//Rumtime:83msclass Solution {public:    int countPrimes(int n) {        int count = 0;        bool *b = new bool[n];        b[2] = true; //2是偶數，但不能被篩掉，需要特殊考慮        for (int i = 3; i < n; i++)        {            if (i & 1)            {                b[i] = true; //奇數            }            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;    }};