Count Primes

Description:

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

References:

分析：运用Sieve of Eratosthenes方法，首先筛掉2的倍数，然后筛掉3的倍数，5的倍数，7的倍数等。注意第二层循环的起点为i而不是1。运行时间312ms。想想该如何优化吧

 1 class Solution {
 2 public:
 3     int countPrimes(int n) {
 4         if(n < 2) return 0;
 5         
 6         vector<bool> result(n, true);
 7         for(int i = 2; i < sqrt(n); i++){
 8             if(result[i]){
 9                 for(int j = i; i * j < n; j++)
10                     result[i*j] = false;
11             }
12         }
13         int count = 0;
14         for(int i = 2; i < n; i++){
15             if(result[i]) count++;
16         }
17         return count;
18     }
19 };