prime=[True]*n prime[:2]=[False,False] forbaseinxrange(2,int((n-1)**0.5)+1): ifprime[base]: prime[base**2::base]=[False]*len(prime[base**2::base]) returnsum(prime) 算法都是一样的,细节处有些不同,比方在对开头if的判断;对prime数组赋值的语句。这里我很好奇,这些改变到底有没有...
Leetcode Count Prime Description: Count the number of prime numbers less than a non-negative number,n Hint:The number n could be in the order of 100,000 to 5,000,000. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 #define NO...
Count the number of prime numbers less than a non-negative number, n. 基本思路:筛法 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 count...
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...
<LeetCode OJ> 204. Count Primes Count the number of prime numbers less than a non-negative number, n. 分析: 思路首先:一个数不是合数就是素数,合数更好推断呢! 合数:不论什么一个合数都能够表现为适当个素数的乘积的形式, 所以我们仅仅用小于sqrt(number)的素数去除要推断的数就可以,...
LeetCode 204 Count Primes Problem: Count the number of prime numbers less than a non-negative number, n. Summary: 判断小于某非负数n的质数个数. Solution: 用所谓的"刷质数表"的方式先用HashTable记录小于n的所有数是质数还是合数,再逐一数出. 看了题目中的Hint才知道这种方法有一个更加高大上的名字...
2716-prime-subtraction-operation 2727-number-of-senior-citizens 2755-extra-characters-in-a-string 2778-frequency-tracker 2792-neighboring-bitwise-xor 2794-maximum-number-of-moves-in-a-grid 2800-minimum-string-length-after-removing-substrings 2803-modify-graph-edge-weights 2855-maximum-number-of-jumps...
A set of practice note, solution, complexity analysis and test bench to leetcode problem set - leetcode/CountSubArrayFixBound.drawio at b58bcceb0ea27d0756ad72fb6a64b3b547fae221 · brianchiang-tw/leetcode
prime= [1]*nforiinrange(2,int(n*0.5)+1): prime[i*i:n:i] = [0] * len(prime[i*i:n:i])returnsum(prime)-2#-2 because 0 and 1 is not a prime number 更快优化 importmathclassSolution(object):defcountPrimes(self, n):""":type n: int ...
2. use non-prime numbers as the step 3. use i*i as the start. 4. use count-- in every loop, avoiding another traversal. 1defcountPrime(n):2ifn<=2:3return04ifn==3:5return16prime=[True]*n7count=n-28rt=sqrt(n)9foriinrange(2,rt+1):10ifprime[i]==True:11forjinxrange(i*i...