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…

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…

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[]…

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. 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…

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…

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…

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…

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 +…