0204. Count Primes (E)

Count Primes (E)

题目

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

Example:

Input: 10
Output: 4
Explanation: There are 4 prime numbers less than 10, they are 2, 3, 5, 7.

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题意

计算小于指定整数n 的素数的个数。

思路

参考两种素数筛：埃氏筛和欧拉筛

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代码实现

Java

埃氏筛

class Solution {
    public int countPrimes(int n) {
        boolean[] p = new boolean[n];
        int count = 0;
        for (int i = 2; i < n; i++) {
            if (p[i] == false) {
                count++;
                for (int j = 2 * i; j < n; j += i) {
                    p[j] = true;
                }
            }
        }
        
        return count;
    }
}

欧拉筛

class Solution {
    public int countPrimes(int n) {
        boolean[] p = new boolean[n];
        int[] primes = new int[n];
        int count = 0;

        for (int i = 2; i < n; i++) {
            if (p[i] == false) {
                primes[count++] = i;
            }
            for (int j = 0; j < count; j++) {
                if (i * primes[j] >= n) {
                    break;
                }
                p[i * primes[j]] = true;
                if (i % primes[j] == 0) {
                    break;
                }
            }
        }

        return count;
    }
}

JavaScript

/**
 * @param {number} n
 * @return {number}
 */
var countPrimes = function (n) {
  const sieve = []
  const primes = []

  for (let i = 2; i < n; i++) {
    if (!sieve[i]) {
      primes.push(i)
    }

    for (let j = 0; j < primes.length; j++) {
      if (i * primes[j] >= n) break
      sieve[i * primes[j]] = true
      if (i % primes[j] === 0) break
    }
  }

  return primes.length
}