Leetcode - Count Primes

最新推荐文章于 2022-09-27 11:10:45 发布

原创最新推荐文章于 2022-09-27 11:10:45 发布 · 129 阅读

0 ·

CC 4.0 BY-SA版权

Math 专栏收录该内容

22 篇文章

订阅专栏

本文详细介绍了如何使用埃拉托斯特尼筛法来高效计算指定范围内质数的数量，并提供了一个C语言实现的示例，包括一些优化技巧以减少时间和空间复杂度。

[ref]
[url]https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes[/url]
[url]https://leetcode.com/discuss/34622/my-c-solutions-in-44ms-time-nearly-o-n-and-space-nearly-o-n[/url]


public class Solution {
    // https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
    // https://leetcode.com/discuss/34622/my-c-solutions-in-44ms-time-nearly-o-n-and-space-nearly-o-n
    public int countPrimes(int n) {
        if (n < 3) return 0;
        int sqrtN = (int)Math.sqrt(n); // trick 0: use sqrt
        int counter = n / 2; //trick 1: all even num except 2 is not prime 
        boolean[] prime = new boolean[n];
        for (int i = 1; i < n; i += 2)
            prime[i] = true;
        for (int i = 3; i <= sqrtN; i += 2) { // trick 2: skip even num
            if (prime[i]) {
                // mark multiples of i not prime
                for (int j = i * i; j < n; j += i) { // trick 3: start from i*i
                    if (prime[j]) {
                        prime[j] = false;
                        counter--; // trick 4: avoid another loop
                    }
                }
            }
        }
        return counter;
    }
}