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...
1classSolution {2public:3intcountPrimes(intn) {4bool*tag =newbool[n];5int*prime =newint[n];6intcnt =0;7memset(tag,true,sizeof(bool) *n);8for(inti =2; i < n; ++i) {9if(tag[i]) prime[cnt++] =i;10for(intj =0; j < cnt && i * prime[j] < n; ++j) {11tag[i ...
<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...
The most important part of this solution is sorting all entries. SinceMap.Entrydoesn't implement theComparableinterface, we need to write our own custom Comparator to sort the entries. If you look at my implementation, I am comparing entries on their values because that's what we want. Many...
The runtime complexity ofisPrimefunction would be O(n) and hence counting the total prime numbers up tonwould be O(n2). Could we do better? Solution: (筛选法) 1. use square root of n. 2. use non-prime numbers as the step
classSolution{public://方法一:逐个判断O(n^(3/2))算法,TLEintcountPrimes1(intn){intcount =0;for(inti =2; i < n; ++i) {if(isPrime(i)) ++count; }returncount; }boolisPrime(intn){for(inti =2; i <=sqrt(n); ++i)if(n % i ==0)returnfalse;returntrue; ...
is_prime[0], is_prime[1] =False, Falseforxinrange(2, int(n ** 0.5) + 1):ifis_prime[x]: p= x *xwhilep <n: is_prime[p]=False p+=xreturnsum(is_prime) Java解法 publicclassSolution { public int countPrimes(int n) { ...
两个方法类似,感觉方法二 循环的次数少了些。 class Solution: def countPrimes(self, n: int) -> int: primes = [1] * n for i in range(2, int(n**0.5)+1): if primes[i]: primes[i*i:n:i] = [0] * ((n-1-i*i)//i+1) return max(0, sum(primes) - 2)发布...