Description:

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

一刷
題解：這題不用brute-force來求解。
i從2到n遍歷, j從2到i遍歷。ij<n, 用boolean數組，notPrime[Ij] = true

public class Solution {
    public int countPrimes(int n) {
        boolean[] notPrime = new boolean[n];
        int count = 0;
        for(int i=2; i<n; i++){
            if(notPrime[i] == false){
                count++;
                for(int j=2; i*j<n; j++){
                    notPrime[i*j] = true;
                }
            }
        }
        return count;
    }
}