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

posted on 2015-04-29 10:04  Step-BY-Step  阅读(316)  评论(0)  编辑  收藏  举报

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