LeetCode： Count Primes（计算n以内素数个数：高效算法）

LeetCode： Count Primes（计算n以内素数个数：高效算法）

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Description:

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

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A prime number is a natural number that has exactly two distinct natural number divisors: 1 and itself.
https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes#Algorithm_complexity
埃拉托色尼筛选法：Sieve of Eratosthenes。
时间复杂度：O(n log log n)。
下面是图示过程，还有leetcode上hint用java编写的。

class Solution {
public:
    int countPrimes(int n) {
        if(!n||n==1)  return 0;
        vector<bool> isPrime(n,true);
        // Loop's ending condition is i * i < n instead of i < sqrt(n)
        // to avoid repeatedly calling an expensive function sqrt().
        for(int i=2;i*iif(!isPrime[i]) continue;
            //填表起点i*i，如3*3，因为3*2已填，步长+i
            for(int j=i*i;jfalse;
            }
        }
        int count=0;
        for(int i=2;iif(isPrime[i])  ++count;
        }
        return count;
    }
};

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别人代码用的i*i与n比较，值得学习： Loop’s ending condition is i * i < n instead of i < sqrt(n) to avoid repeatedly calling an expensive function sqrt().