LeetCode Count Primes

原题链接在这里：https://leetcode.com/problems/count-primes/

题目：

Description:

Count the number of prime numbers less than a non-negative number, n.

题解：

网上查查，原来有一种方法叫：Sieve of Eratosthenes 的方法。时间复杂度为O(nloglogn),空间复杂度为O(n).

AC Java:

public class Solution {
public int countPrimes(int n) {
if(n < 3){
return 0;
}
boolean [] isPrime = new boolean
;
Arrays.fill(isPrime, true);
int res = n-2;
for(int i = 2; i*i<n; i++){
if(isPrime[i]){
for(int j = i; i*j<n; j++){
if(isPrime[i*j]){
isPrime[i*j] = false;
res--;
}
}
}
}
return res;
}
}

跟上Perfect Squares