knight3396的专栏

Description:Count the number of prime numbers less than a non-negative number, n.求小于n的素数的数量。 最传统的做法就是做一个判定函数，从2开始到n判断，看有多少素数。判定素数的方法就是从2到根号n，看看是不是可以被n整除。但是看了提示后，发现还有别的便捷的方法来解，就是原始的Sieve of Eratosthene

Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers in this range, inclusive.For example, given the range [5, 7], you should return 4.

There are a total of n courses you have to take, labeled from 0 to n - 1.Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair:

Implement a trie with insert, search, and startsWith methods.Note: You may assume that all inputs are consist of lowercase letters a-z.这个题比较简单，就是实现一个单词树，单词树的相关资料可以参考：http://www.cppblog.com/yuyang7/arc