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...
1classSolution {2public:3intcountPrimes(intn) {4vector<bool> p(n,true);5//去掉2以外的所有偶数6for(inti =4; i < n; i +=2) p[i] =false;7//上一步已经去掉了偶数,所以这里可以使用i += 2,j += 28for(inti =3; i * i < n; i +=2) {9if(p[i])for(intj =3; i * j ...
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...
}/** * 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 ...
技术标签: Leetcode文章作者:Tyan 博客:noahsnail.com | CSDN | 简书 1. Description 2. Solution Version 1 class Solution { public: int countPrimes(int n) { if(n < 3) { return 0; } int count = 1; for(int i = 3; i < n; i += 2) { if(isPrime(i)) { count++; } } ...
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++) { ...
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Debugging... Run Submit Description Description Editorial Editorial Solutions Solutions Submissions Code Testcase Test Result Test Result 🔥 Join LeetCode to Code! View your Submission records here Register or Sign In Ln 1, Col 1 Case 1
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); ...