[LeetCode] 204. Count Primes

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

Example 1:

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

Example 2:

Input: n = 0
Output: 0

Example 3:

Input: n = 1
Output: 0

Constraints:

• 0 <= n <= 5 * 106

计数质数。

题意是找出小于N范围内的整数里面所有的质数（prime number）的个数。思路是打表法，在2 ~ N范围内，先找出所有的质数，同时找出这些质数的倍数，标记为false。这样剩下没有被标记到的数字就都是质数了。

时间O(n)

空间O(n)

Java实现

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

 

JavaScript实现

/**
 * @param {number} n
 * @return {number}
 */
var countPrimes = function(n) {
    let notPrimes = new Array(n);
    for (let i = 0; i < n; i++) {
        notPrimes[i] = false;
    }
    let res = 0;
    for (let i = 2; i < n; i++) {
        if (notPrimes[i] === false) {
            res++;
            for (let j = 2; i * j < n; j++) {
                notPrimes[i * j] = true;
            }
        }
    }
    return res;
};

 

LeetCode 题目总结