LeetCode OJ：Count Primes（质数计数）

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

计算小于n的质数的个数，当然就要用到大名鼎鼎的筛法了，代码如下，写的有点乱不好意思。

 class Solution {
 public:
     int countPrimes(int n) {
         vector<int> vtor(n + , );
         vector<int> ret;
         for (int i = ; i <= n; ++i)
             vtor[i] = i;
         for (int i = ; i < n; ++i){//边界条件注意
             if (vtor[i] != -){
                 ret.push_back(vtor[i]);
                 int tmpPrime = vtor[i];
                 for (int mul = ; tmpPrime * mul <= n; mul++){
                     vtor[tmpPrime * mul] = -;
                 }
             }
         }
         return ret.size();
     }
 };

java版本的代码如下所示：

public class Solution {
    public int countPrimes(int n) {
        int [] prime = new int[n+]; //这里其实用boolean更好，懒得改了
        int count = ;
        for(int i = ; i < n+; ++i){
            prime[i] = i;
        }
        for(int i = ; i < n+; ++i){
            if(prime[i] == -)
                continue;
            else{
                count++;
                for(int mul = ; mul * i < n+; ++mul){
                    prime[mul*i] = -;
                }
            }
        }
        return count;
    }
}