LeetCode OJ：Count Primes（质数计数）-优快云博客

本文介绍了使用筛法计算小于指定非负整数n的质数个数的方法，并提供了Java版代码实现。通过迭代筛选出所有质数，优化边界条件处理。

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。

 1 class Solution {
 2 public:
 3     int countPrimes(int n) {
 4         vector<int> vtor(n + 1, 0);
 5         vector<int> ret;
 6         for (int i = 0; i <= n; ++i)
 7             vtor[i] = i;
 8         for (int i = 2; i < n; ++i){//边界条件注意
 9             if (vtor[i] != -1){
10                 ret.push_back(vtor[i]);
11                 int tmpPrime = vtor[i];
12                 for (int mul = 1; tmpPrime * mul <= n; mul++){
13                     vtor[tmpPrime * mul] = -1;
14                 }
15             }
16         }
17         return ret.size();
18     }
19 };

java版本的代码如下所示：

public class Solution {
    public int countPrimes(int n) {
        int [] prime = new int[n+1]; //这里其实用boolean更好，懒得改了
        int count = 0;
        for(int i = 0; i < n+1; ++i){
            prime[i] = i;
        }
        for(int i = 2; i < n+1; ++i){
            if(prime[i] == -1)
                continue;
            else{
                count++;
                for(int mul = 2; mul * i < n+1; ++mul){
                    prime[mul*i] = -1;
                }
            }
        }
        return count;
    }
}