204. Count Primes

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本文介绍了一种计算小于非负整数n的所有素数数量的方法。通过标记法筛选非素数，逐步迭代并跳过已知的合数，提高了算法效率。

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

大致思路差不多答案要简洁一些注意结束条件不是 i<num/2 而是i*i<=num

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;
}

it starts from 2, the first prime, then mark the multip of 2 as true in notPrime, so the loop of i will skip them. the next prime is 3, do the same thing. Then it is 4, which 2*2, so the not prime is true, and will skip to next.