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…

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…

计算所有小于非负整数 n 的质数数量. 详见:https://leetcode.com/problems/count-primes/description/ Java实现: 埃拉托斯特尼筛法: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,一次类推,直到根号n,此时数组中未被标记的数字就是质数. class Solution { public int countPrimes(int n) { int res=0; boolean[]…

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…

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.…

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…

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(…

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…

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…