leetcode-204. Count Primes

Description:

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

Credits:

Special thanks to @mithmatt for adding this problem and >creating all test cases.

Hint:

Let’s start with a isPrime function. To determine if a >number is prime, we need to check if it is not divisible by >any number less than n. The runtime complexity of >isPrime function would be O(n) and hence counting the >total prime numbers up to n would be O(n2). Could we >do better?

As we know the number must not be divisible by any >number > n / 2, we can immediately cut the total >iterations half by dividing only up to n / 2. Could we still >do better?

思路：如果用传统的从下往上遍历依次判断，数字大了显然不行。改进方法：空间换时间，给n个数设标志位，从小的开始依次乘积改标志位。直接看代码吧：

class Solution {
//传统的遍历判断是否是素数效率太低了，只能用空间换时间
public:
int countPrimes(int n) {
vector<bool> isPrime(n+1,true);
int primeNum = 0;
int upperBound = sqrt(n);
for(int i=2;i<n;i++)
{
if(isPrime[i])
{
primeNum++;

if(i > upperBound)
{
continue;
}
//小于n的能组合分解的都为合数
for(int j = i;i*j < n;j++)
{
isPrime[i*j] = false;
}
}
}
return primeNum;
}
};