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统计所有小于非负整数 n 的质数的数量。
示例 1:
输入:n = 10 输出:4 解释:小于 10 的质数一共有 4 个, 它们是 2, 3, 5, 7 。
示例 2:
输入:n = 0 输出:0
示例 3:
输入:n = 1 输出:0
提示:
来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/count-primes 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
const countPrimes = n => { let nums = [] for (let i = 2; i < n; i++) { let flag = true for (let num of nums) { let res = i / num if (res === ~~res) { flag = false break } } if (flag) nums.push(i) } return nums.length }
const countPrimes = n => { let arr = [], count = 0 for (let i = 2; i < n; i++) { if (!arr[i]) { for (let j = i * i; j < n; j += i) { arr[j] = true } count++ } } return count }
The text was updated successfully, but these errors were encountered:
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统计所有小于非负整数 n 的质数的数量。
示例 1:
示例 2:
示例 3:
提示:
The text was updated successfully, but these errors were encountered: