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...
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[...
02 第一种解法 判断一个数n是否为素数,就需要判断1到n-1之间的数能否被n整除,能够被整除说明不是素数,否则就是素数。 publicintcountPrimes(intn){if(n <=2) {return0; }intcount=0;for(inti=2;i<n; i++) {booleanflag=true;for(intj=2; j<i; j++) {if(i%j ==0) { flag =false; } ...
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 ...
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]; ...
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...
【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...
【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) ...