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.…

Write a program to find the nth super ugly number. Super ugly numbers are positive numbers whose all prime factors are in the given prime list primes of size k. Example: Input: n = 12, primes = [2,7,13,19] Output: 32 Explanation: [1,2,4,7,8,13,14,16,…

Write a program to find the nth super ugly number. Super ugly numbers are positive numbers whose all prime factors are in the given prime list primes of size k. For example, [1, 2, 4, 7, 8, 13, 14, 16, 19, 26, 28, 32] is the sequence of the first 12…

题目: Write a program to find the nth super ugly number. Super ugly numbers are positive numbers whose all prime factors are in the given prime list primes of size k. For example, [1, 2, 4, 7, 8, 13, 14, 16, 19, 26, 28, 32] is the sequence of the first…

264. 丑数 II 题目描述 编写一个程序,找出第 n 个丑数. 丑数就是只包含质因数 2, 3, 5 的正整数. 示例: 输入: n = 10 输出: 12 解释: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 是前 10 个丑数. 说明:   1. 1 是丑数. 2. n 不超过1690. 想法 三指针法.一部分是丑数数组,另一部分是权重2,3,5.下一个丑数,定义为丑数数组中的数乘以权重,所得的最小值. 那么,2该乘以谁?3该乘以谁?5该乘以谁? 其一,使用三个指针idx…

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…

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(…

▶ 三个与丑数相关的问题 ▶ 第 263题,判定一个数字是否是丑数,即其素因子是否仅由 2,3,5 构成. ● 常规消除判别,4 ms class Solution { public: bool isUgly(int num) { ) return false; ); num /= ); ); num /= ); ); num /= ); ; } }; ● 递归方法,6 ms class Solution { public: bool isUgly(int num) { ) return fal…

丑数系列的题看这一道就可以了 /* 和ugly number2差不多,不过这次的质因子多了,所以用数组来表示质因子的target坐标 target坐标指的是这个质因子此次要乘的前任丑数是谁 */ public int nthSuperUglyNumber(int n, int[] primes) { //记录相乘坐标,存的是每个质因子对应相乘搭档在丑数数组中的下标 int[] target = new int[primes.length]; //动态规划数组 int[] dp = new int…

题目如下: 解题思路:总结一下这么几点,一出一进,优先级队列排序,保证每次输出的都是当前的最小值.解法大致如图: 代码如下: #include<map> #include<queue> using namespace std; struct node { node(int k,int v) { key = k; val = v; } friend bool operator< (node n1, node n2) { return n1.val > n2.val; }…

编写一段程序来寻找第 n 个超级丑数.超级丑数是指其所有质因数都在长度为k的质数列表primes中的正整数.例如,[1, 2, 4, 7, 8, 13, 14, 16, 19, 26, 28, 32],是给定长度为 4 的质数列表primes = [2, 7, 13, 19]的前 12 个超级丑数.注意:(1) 1是任何给定的primes的超级丑数.(2) 给定primes中的数字以升序排列.(3) 0 < k ≤ 100, 0 < n ≤ 106, 0 < primes[i] <…

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 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…

Description: Count the number of prime numbers less than a non-negative number, n. 题目标签:Hash Table 题目给了我们一个n, 让我们找出比n小的 质数的数量. 因为这道题目有时间限定,不能用常规的方法. 首先建立一个boolean[] nums,里面初始值都是false,利用index 和 boolean 的对应,把所有prime number = false:non-prime number = tr…

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…

Problem: Count the number of prime numbers less than a non-negative number, n. Summary: 判断小于某非负数n的质数个数. Solution: 用所谓的"刷质数表"的方式先用HashTable记录小于n的所有数是质数还是合数,再逐一数出. 看了题目中的Hint才知道这种方法有一个更加高大上的名字Sieve of Eratosthenes 即在每判断一个数为质数时,将它的bei'shu倍数均计为合数. c…

Description: Count the number of prime numbers less than a non-negative number, n. 解题思路: 空间换时间,开一个空间为n的数组,因为非素数至少可以分解为一个素数,因此遇到素数的时候,将其有限倍置为非素数,这样动态遍历+构造下来,没有被设置的就是素数. public int countPrimes(int n) { if (n <= 2) return 0; boolean[] notPrime = new boo…

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…

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…

题目描述: Description: Count the number of prime numbers less than a non-negative number, n. 解题思路: 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 comp…

Count the number of prime numbers less than a non-negative number, n. 问题:找出所有小于 n 的素数. 题目很简洁,但是算法实现的优化层次有很多层.其中最主要思想的是采用 Sieve of Eratosthenes 算法来解答. 大思路为: 找出 n 范围内所有合数,并做标记. 未做标记的即为素数,统计未做标记的数个数即为原题目解. 如何找到 n 范围内所有合数? 将第一个素数 2 赋值给 i. 当 i 小于 n 时:(2)…

原题 原题链接 Description: Count the number of prime numbers less than a non-negative number, n. 计算小于非负数n的素数个数. 思路 这题用埃拉托斯特尼筛法来做效果比较好,普通的方法基本会TLE.但是在用了埃拉托斯特尼筛法之后,还有一些细节值得注意: (1)首先我们该用 i*i<=n 替代 i<=sqrt(n) 来避免使用 sqrt() ,因为sqrt()的操作是比较expensive的. (2)当要表示两个状…

Description: Count the number of prime numbers less than a non-negative number, n. 题解:就是线性筛素数的模板题. class Solution { public: int countPrimes(int n) { ; vector<,); ;i<n;i++){ if(is_prime[i]){ ans++; *i;j<n;j+=i){ is_prime[j]=; } } } return ans; } }…

题目: 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. 分析: 统计所有小于非负整数 n 的质数的数量. 这里使用埃拉托斯特尼筛法.要得到自然数n以内的全部素数,必须把不大于√n的所有素数的倍数剔除,剩…

#-*- 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…

题意:统计小于n的质数个数. 作为一个无节操的楼主,表示用了素数筛法,并没有用线性素数筛法. 是的,素数筛法并不是该题最佳的解法,线性素数筛法才是. 至于什么是素数筛法,请百度吧. class Solution { public: int countPrimes(int n) { bool *isp= new bool[n]; ; i < n; ++i) { isp[i]= true; } ; i * i< n; ++i)//素数筛法 { if(isp[i]){ for(int j = i +…

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…

https://leetcode.com/problems/ugly-number/ Write a program to check whether a given number is an ugly number. Ugly numbers are positive numbers whose prime factors only include 2, 3, 5. For example, 6, 8 are ugly while 14 is not ugly since it include…

Write a program to check whether a given number is an ugly number. Ugly numbers are positive numbers whose prime factors only include 2, 3, 5. Example 1: Input: 6 Output: true Explanation: 6 = 2 × 3 Example 2: Input: 8 Output: true Explanation: 8 = 2…