Description:
Count the number of prime numbers less than a non-negative number, n.
https://leetcode.com/problems/count-primes/
找出所有小于n的数中的质数。
删数法。开一个1到n的数组，删除所有2的倍数，3的倍数...直到√n的倍数，最后剩下的就是质数。
1 /** 2 * @param {number} n 3 * @return {number} 4 */ 5 var countPrimes = function(n) { 6 var count = 0, i, j, dict = [], 7 len = Math.floor(Math.sqrt(n)); 8 for(i = 2; i <= len; i++) 9 for(j = 2; i * j <= n; j++) 10 dict[i * j] = false; 11 for(i = 2; i < n; i++) 12 if(dict[i] === undefined) count++; 13 return count; 14 };
[LeetCode][JavaScript]Count Primes
原文：http://www.cnblogs.com/Liok3187/p/5224903.html