Problem:

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

Idea:

Sieve of Eratosthenes

Time Complexity:

O(nloglogn)

Space Complexity:

O(n)

Solution:

C++

// Author: Huahua
class Solution {
public:
    int countPrimes(int n) {
        if (n < 3) return 0;
        
        vector<unsigned char> f(n, 1);
        f[0] = f[1] = 0;
        for(long i=2;i<=sqrt(n);++i) {
            if(!f[i]) continue;
            for(long j=i*i;j<n;j+=i)
                f[j] = 0;
        }
        
        int ans = accumulate(f.begin(), f.end(), 0);
        return ans;
    }
};

Java

// Author: Huahua
// Running time: 16 ms
class Solution {
  public int countPrimes(int n) {
    int ans = 0;
    boolean[] isPrime = new boolean[n + 1];
    Arrays.fill(isPrime, true);
    isPrime[0] = false;
    if (n > 0) isPrime[1] = false;
    for (int i = 2; i < n; ++i) {
      if (!isPrime[i]) continue;
      ++ans;
      for (int j = 2 * i; j < n; j += i)
        isPrime[j] = false;
    }
    return ans;
  }
}

Python3

"""
Author: Huahua
Running time: 800 ms
"""
class Solution:
  def countPrimes(self, n):
    if n < 3: return 0
    isPrime = [True] * (n + 1)
    isPrime[0] = False
    isPrime[1] = False
    ans = 0
    for i in range(2, n):
      if not isPrime[i]: continue
      ans += 1
      for j in range(2 * i, n, i):
        isPrime[j] = False
    return ans

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