Count the number of prime numbers less than a non-negative number,n. 找出小于n的素数个数。 1、用最淳朴的算法果然超时了。 publicclassSolution {publicintcountPrimes(intn) {if(n < 2){return0; }intresult = 0;for(inti = 2; i < n; i++){if(isPrimes(i)){ result++; } }returnresult...
204. Count Primes Count the number of prime numbers less than a non-negative number,n. 题意:计算小于非负数的质数数。 1publicclassSolution {2publicintcountPrimes(intn) {3//能够在这里直接对动态数组进行初始化4int[] mask=newint[n];5intcount=0;6for(inti=2;i<n;i++)7{8if(mask[i]==...
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 ...
25. LeetCode题解, 大循环执行sqrt(n)次, 就能将notPrime填写完整! 不过这样就不能在大循环中计数了, 需要单独写个循环遍历notPrime用来统计质数的个数 public int countPrimes(int n) { if(n <=1 ) return 0; boolean[] notPrime = new boolean[n]; notPrime[0] = true; notPrime[...
[LeetCode] Count Primes Problem Count the number of prime numbers less than a non-negative number, n. Note 用数组flag标记非质数,每当出现一个flag[i]为false,计数器count加一。 关于质数有三点: 大于3的质数一定是奇数,如3,5,7; 奇数中的非质数也一定是奇数的乘积。
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...
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;...
Runtime: 11 ms, faster than 94.69% of Java online submissions for Count Primes. 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]; ...
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...
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; ...