leetcode-easy-math-204 Count Primes-NO】的更多相关文章

#-*- coding: UTF-8 -*- #Hint1:#数字i,i的倍数一定不是质数,因此去掉i的倍数,例如5,5*1,5*2,5*3,5*4,5*5都不是质数,应该去掉#5*1,5*2,5*3,5*4 在数字1,2,3,4的时候都已经剔除过,因此数字5,应该从5*5开始#Hint2:#例:#2 × 6 = 12#3 × 4 = 12#4 × 3 = 12#6 × 2 = 12#显然4 × 3 = 12,和6 × 2 = 12不应该分析,因为在前两式中已经知道12不是质数,因此如果数字i是…
Description: Count the number of prime numbers less than a non-negative number, n. 分析: 思路首先:一个数不是合数就是素数,合数更好推断呢! 合数:不论什么一个合数都能够表现为适当个素数的乘积的形式, 所以我们仅仅用小于sqrt(number)的素数去除要推断的数就可以, 由于合数在sqrt(number)以内一定有素数的质因子 比方要推断100以内的素数,仅仅需推断合数就可以.仅仅用10以内的2,3.5.7就够…
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.…
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…
Queston 204. Count Primes Solution 题目大意:给一个数,求小于这个数的素数的个数 思路:初始化一个boolean数组,初始设置为true,先遍历将2的倍数设置为false,再遍历3并将3的倍数置为false... Java实现: 此法超时 public int countPrimes(int n) { // 2, 3, 5, 7 boolean[] candidate = new boolean[n]; Arrays.fill(candidate, Boolea…
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 素数筛法 参考资料 日期 [LeetCode] 题目地址:https://leetcode.com/problems/count-primes/ Total Accepted: 36655 Total Submissions: 172606 Difficulty: Easy 题目描述 Count the number of prime numbers…
Count Primes 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. 解法1 埃氏筛法 class Solution { public: int countPrimes(int n) { if(…
Description: Count the number of prime numbers less than a non-negative number, n. Credits:Special thanks to @mithmatt for adding this problem and creating all test cases. 解析:大于1的自然数,该自然数能被1和它本身整除,那么该自然数称为素数. 方法一:暴力破解,时间复杂度为O(N^2) 代码如下: public class…
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…
Description:Count the number of prime numbers less than a non-negative number, n. Hint: Let's start with a isPrime function. To determine if a number is prime, we need to check if it is not divisible by any number less than n. The runtime complexity…