LeetCode - Count Primes
判断小于n数中素数的个数,如果用普通的判断方法会超时,这里使用筛选法。
具体请参考:http://blog.csdn.net/liukehua123/article/details/5482854
public class Solution { public int countPrimes(int n) { if(n <= 2) return 0; int count = 0; boolean[] prime = new boolean[n+1]; for(int i=1; i<n+1; i++) { if(i % 2 == 0) { prime[i] = false; } else { prime[i] = true; } } for(int i=3; i<=Math.sqrt(n); i++) { if(prime[i]) { for(int j=2*i; j<n; j+=i) { prime[j] = false; } } } for(int i=1; i<n; i++) { if(prime[i]) { count++; } } return count; } }
作者:Pickle
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