Count Primes

Description:

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

 

 1 public class Solution {

 2     public int countPrimes(int n) {

 3         if(n <= 2) return 0;

 4         boolean[] arr = new boolean[n];//use false to represent prime numbers

 5         //Actually here only use 1 ~ n-1

 6         int sqr = (int)Math.sqrt(n - 1);

 7         for(int i = 2; i <= sqr; i ++){

 8             if(!arr[i]){

 9                 for(int j = i * i; j < n; j += i){

10                     arr[j] = true;

11                 }

12             }

13         }

14         int count = 0;

15         for(int i = 2; i < n; i ++){

16             if(!arr[i]) count ++;

17         }

18         return count;

19     }

20 }

 

Another Solution:

 1 public class Solution {

 2     public int countPrimes(int n) {

 3         if(n < 3) return 0;

 4         HashSet<Integer> primes = new HashSet<Integer> ();

 5         for(int i = 2; i < n; i ++){

 6             if(primes.isEmpty()){

 7                 primes.add(i);

 8             }else{

 9                 Iterator it = primes.iterator();

10                 boolean isPrime = true;

11                 while(it.hasNext()){

12                     Integer tmp = (Integer) it.next();

13                     if(i % tmp == 0) isPrime = false;

14                 }

15                 if(isPrime) primes.add(i);

16             }

17         }

18         return primes.size();

19     }

20 }

This one return Time Limit Exceeded for case: 499979