Description:

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

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How Many Primes Are There?

Sieve of Eratosthenes

Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.

計數出小於非負整數n的質數數量。質數（prime number）又稱素數，有無限個。質數定義為在大於1的自然數中，除了1和它本身以外不再有其他因數。

解法：埃拉托斯特尼篩法 Sieve of Eratosthenes

如果一個數是另一個數的倍數，那這個數肯定不是質數。利用這個性質，可以建立一個質數數組，從2開始將素數的倍數都標註為不是質數。第一輪將4、6、8等表為非質數，然後遍歷到3，發現3沒有被標記為非質數，則將6、9、12等標記為非質數，一直到N為止，再數一遍質數數組中有多少質數。

Java:

public class Solution {
    public int countPrimes(int n) {
        boolean[] prime = new boolean[n];
        Arrays.fill(prime, true);
        for(int i = 2; i < n; i++){
            if(prime[i]){
                // 將i的2倍、3倍、4倍...都標記為非素數
                for(int j = i * 2; j < n; j =  j + i){
                    prime[j] = false;
                }
            }
        }
        int count = 0;
        for(int i = 2; i < n; i++){
            if(prime[i]) count++;
        }
        return count;
    }
}

Python:

class Solution:
    # @param {integer} n
    # @return {integer}
    def countPrimes(self, n):
        isPrime = [True] * max(n, 2)
        isPrime[0], isPrime[1] = False, False
        x = 2
        while x * x < n:
            if isPrime[x]:
                p = x * x
                while p < n:
                    isPrime[p] = False
                    p += x
            x += 1
        return sum(isPrime)

Python:

class Solution(object):
    def countPrimes(self, n):
        """
        :type n: int
        :rtype: int
        """
        if n <= 2: return 0
        vis = [False] * n
        for i in range(2, int(n ** 0.5) + 1):
            if vis[i]: continue
            j = i
            while j * i < n:
                vis[j * i] = True
                j += 1
        ans = 0
        for i in range(2, n):
            if not vis[i]: ans += 1
        return ans

C++:

class Solution {
public:
    int countPrimes(int n) {
        if(!n||n==1)  return 0;
        vector<bool> isPrime(n,true);
        // Loop‘s ending condition is i * i < n instead of i < sqrt(n)
        // to avoid repeatedly calling an expensive function sqrt().
        for(int i=2;i*i<n;++i)
        {
            if(!isPrime[i]) continue;
            //填表起點i*i，如3*3，因為3*2已填，步長+i
            for(int j=i*i;j<n;j+=i)
            {
                isPrime[j]=false;
            }
        }
        int count=0;
        for(int i=2;i<n;++i)
        {
            if(isPrime[i])  ++count;
        }
        return count;
    }
};

[LeetCode] 204. Count Primes 計數質數