classSolution:defcountPrimes(self, n: int) ->int: count,flag= 0,[True]*(n)#1代表质数,0代表非质数foriinrange(2,n):#从2开始遍历,i代表当前第一个数,i代表这个数所在位置ifflag[i]:#如果当前位置判定为True的话count += 1forjinrange(i*i,n,i):#将该质数的所有倍数都设定为False,即非质数f...
primes=[True]*n primes[0]=primes[1]=False foriinrange(2,int(n**0.5)+1): ifprimes[i]: primes[i*i: n: i]=[False]*len(primes[i*i: n: i]) returnsum(primes) 下面的code是有人针对上面这个code进行改进的code: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 classSolution(object):...
classSolution(object):defcountPrimes(self,n):""":type n: int:rtype: int"""ifn<=2:return0prime=[True]*ni=3sqrtn=pow(n,0.5)count=n//2# 偶数一定不是质数whilei<=sqrtn:j=i*iwhilej<n:ifprime[j]:count-=1prime[j]=Falsej+=2*ii+=2whilei<=sqrtnandnotprime[i]:i+=2returncount...
import java.util.Arrays; public class Solution3 { public int countPrimes(int n) { boolean[] primes = new boolean[n]; Arrays.fill(primes, true); for (int i = 2; i < n; i++) { // 每一轮第一个没有被划去的数肯定是质数 if (primes[i]) { for (int j = i + i; j < n;...
2, 寻找到下一个未被筛除的数。如3. 再筛掉以3为因子的数。 3, 反复步骤2. 时间复杂度为O(n) class Solution { public: int countPrimes(int n) { vector<int> sieve(n, true); int count = 0; for (int i=2; i<n; i++) {
classSolution {public:intcountPrimes(intn) { vector<bool> num(n -1,true); num[0] =false;intres =0, limit =sqrt(n);for(inti =2; i <= limit; ++i) {if(num[i -1]) {for(intj = i * i; j < n; j +=i) { num[j-1] =false; ...
Memory Usage: 35.9 MB, less than 21.43% of Java online submissions for Count Primes. class Solution { public int countPrimes(int n) { boolean[] notPrimes = new boolean[n]; int count = 0; for (int i = 2; i < n; i++) { ...
【摘要】 Leetcode 题目解析之 Count Primes Description: Count the number of prime numbers less than a non-negative number, n. 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 comple...
Can you solve this real interview question? Count Primes - Given an integer n, return the number of prime numbers that are strictly less than n. Example 1: Input: n = 10 Output: 4 Explanation: There are 4 prime numbers less than 10, they are 2, 3,
【摘要】 这是一道关于素数的LeetCode题目,希望对您有所帮助。 题目概述: Description:Count the number of prime numbers less than a non-negative number, n. 解题方法: 题意是给出n中所有素数的个数。 首先你需要知道判断一个数是不是素数的方法:(最笨方法但有效) ...