Count Primes 解答
Question
Count the number of prime numbers less than a non-negative number, n.
Solution 1
Naive way, to check each number one by one. If a number i is prime, then for x from o to (i - 1), i % x != 0. Time complexity O(n^2).
Solution 2
Sieve of Eratosthenes 算法:
由于一个合数总是可以分解成若干个质数的乘积,那么如果把质数(最初只知道2是质数)的倍数都去掉,那么剩下的就是质数了。
例如要查找100以内的质数,首先2是质数,把2的倍数去掉;此时3没有被去掉,可认为是质数,所以把3的倍数去掉;再到5,再到7,7之后呢,因为8,9,10刚才都被去掉了,而100以内的任意合数肯定都有一个因子小于10(100的开方),所以,去掉,2,3,5,7的倍数后剩下的都是质数了。
1 public class Solution { 2 public int countPrimes(int n) { 3 if (n <= 2) 4 return 0; 5 boolean[] primes = new boolean[n]; 6 primes[0] = false; 7 for (int i = 2; i < n; i++) primes[i] = true; 8 for (int i = 2; i < Math.sqrt(n); i++) { 9 if (primes[i] == true) { 10 int j = 2; 11 while (i * j < n) { 12 primes[i * j] = false; 13 j++; 14 } 15 } 16 } 17 int result = 0; 18 for (int i = 2; i < n; i++) { 19 if (primes[i]) 20 result++; 21 } 22 return result; 23 } 24 }