Description: Count the number of prime numbers less than a non-negative number, n click to show more hints. References: How Many Primes Are There? Sieve of Eratosthenes Credits:Special thanks to @mithmatt for adding this problem and creating all test…

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. References: How Many Primes Are There? Sieve of Eratosthenes Credits:Special…

题意:给一个数n,返回小于n的素数个数. 思路:设数字 k =from 2 to sqrt(n),那么对于每个k,从k2开始,在[2,n)范围内只要是k的倍数的都删掉(也就是说[k,k2)是不用理的,若能被筛掉早就被筛了,保留下来的就是素数).最后统计一下[2,n)内有多少个还存在的,都是素数. 要注意,如果k已经被筛掉了,那么不用再用它来删别人了,因为已经被筛掉,那么现在比k2大的且是k的倍数,都已经被干掉了. class Solution { public: int countPrimes(…

Count Primes Description: Count the number of prime numbers less than a non-negative number, n click to show more hints. References: How Many Primes Are There? Sieve of Eratosthenes Credits: Special thanks to @mithmatt for adding this problem and cre…

问题描述: 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. 思路: 1.最暴力的方法就是循环遍历,用两个for循环嵌套实现,但是整个代码运行时间太长,提交通不过 2.采用'Sieve of Eratos…

质数的判断 埃拉托斯特尼筛法: 算法的过程如下图所示: 我们从2开始遍历到根号n,先找到第一个质数2,然后将其所有的倍数全部标记出来,然后到下一个质数3,标记其所有倍数,依次类推,直到根号n,此时数组中未被标记的数字就是质数. 对于本题,即可采用上述判断质数的方法.…

263. Ugly Number 注意:1.小于等于0都不属于丑数 2.while循环的判断不是num >= 0, 而是能被2 .3.5整除,即能被整除才去除这些数 class Solution { public: bool isUgly(int num) { ) return false; == ) num /= ; == ) num /= ; == ) num /= ; ? true : false; } }; 264. Ugly Number II 用一个数组去存第n个前面的所有整数,然后…

题目大意 https://leetcode.com/problems/count-primes/description/ 204. Count Primes Count the number of prime numbers less than a non-negative number, n. Example: Input: 10Output: 4Explanation: There are 4 prime numbers less than 10, they are 2, 3, 5, 7.…

Count the number of prime numbers less than a non-negative number, n. 计算小于n的质数的个数,当然就要用到大名鼎鼎的筛法了,代码如下,写的有点乱不好意思. class Solution { public: int countPrimes(int n) { vector<, ); vector<int> ret; ; i <= n; ++i) vtor[i] = i; ; i < n; ++i){//边界条件…

examination questions Description: Count the number of prime numbers less than a non-negative number, n References: How Many Primes Are There? Sieve of Eratosthenes Please use the following function to solve the problem: int countPrimes(int n){ } 解题代…