Count Prime

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Count Prime

题目描述：

例子：

具体描述见LeetCode204

解题思路：

主要的思路是我们构建一个bool型的数组，假设全为true。在遍历到第i个数的时候，我们先判断这个数是否是小于sqrt(n)的结果，是的情况下，我们对这个找到的素数的整数倍位置都置为false；如果是true的情况下，在先前的数中已经判断过了所以无需再次判断。

代码如下：

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class Solution {
public:
    int countPrimes(int n) {
    vector<bool> isPrime(n, true);
    
    int primeCount = 0;
    int upperBound = sqrt(n);
    
    for (int i = 2; i < n; i++) {  //Start from 2 since 1 * j == j, all the following numbers will be affected
        if (isPrime[i-1]) {
            primeCount++;
            
            //When i > upper, i(bigger) * j(smaller) has already been checked, where i(bigger) == j(old) and i(smaller) == j(new)
            //Eg. n == 11, 2 * 5 == 10 has been checked when i == 2 and j == 5, so when i == 5 > 3, no need to check 5 * 2 where i == 5 and j == 2.
            if (i > upperBound)
                continue;
                
            for (int j = i; i * j < n; j++) {   //j >= i, so that i(bigger) * j(smaller) has already been checked in previous loops
                isPrime[i*j - 1] = false;
            }
        }
    }
    
    return primeCount;
    }
};