package leetcode func countPrimes(n int) int { isNotPrime := make([]bool, n) for i := 2; i*i < n; i++ { if isNotPrime[i] { continue } for j := i * i; j < n; j = j + i { isNotPrime[j] = true } } count := 0 for i := 2; i < n; i++ { if !is...
}/** * Solution 1: 逐个判断是否是素数,思路简单。但是超时 * Time:O(n^1.5) Space:O(1) */publicintcountPrimes(intn){intcount =0;for(inti =2; i <= n; i++) {if(isPrime(i)) count++; }returncount; } 思路2 // Time: O(n log log n) Space: O(n)publicintcountPrimes(intn){...
leetcode 204 Count Primes 本题使用传统方法,时间复杂度过高,会超时。从discuss学习了两种新的方法。 第一种方法: Sieve of Eratosthenes youtube视频讲解 class Solution { public int countPrimes(int n) { //Java对布尔数组初始化为false boolean[] nf = new boolean[n]; if(n < 3) return 0; int ...
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++) { if (sieve[i]) { ++count; for (int j=i+i; j<n;...
Leetcode 204. Count Primes 算法思路:用质数筛选算法:The Sieve of Eratosthenes ...leetcode - 204. Count Primes Problem: Count the number of prime numbers less than a non-negative number, n. 解释:计算小于n的所有质数。(质数定义为在大于1的自然数中,除了1和它本身以外不再有其他因数。) Solve...
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...
1。 2 为素数。 筛掉以2为因子的数。 即2 * 2, 2*3, 2*4,2*5 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); ...
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;...
链接:https://leetcode-cn.com/problems/count-primes 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。 给出一个正整数n,求(0,n)范围内质数的个数。这道题可以很直观的采用暴力解法,设一个计数器result,当0<= n <= 2时,result肯定为零,因为0 ,1都不是质数。所以只需要从n >= ...