Leetcode 204. Count Primes

public class Solution {
public int countPrimes(int n) {
int cnt = 0;
boolean[] prime = new boolean
;
for (int i=2; i*i<=n; i++)
for (int j=2; i*j<n; j++)
prime[i*j] = true;
for (int k=2; k<n; k++)
if (!prime[k]) cnt++;
return cnt;
}
}

Finding Prime numbers - Sieve of Eratosthenes

O(n^1.5) approach.

public class Solution {
public int countPrimes(int n) {
int count = 0;

for (int i=1; i<n; i++) {
if (isPrime(i)) {
count++;
}
}

return count;
}

public boolean isPrime(int num) {
if (num == 1) {
return false;
}

if (num == 2) {
return true;
}

// Check if num can be divided by even numbers
if (num % 2 == 0) {
return false;
}

// Check if num can be divided by odd numbers
// If num can be divided by p, that num = p * q and assume p <= q,
// we can derive that p <= num^0.5 b/c (p * q) ^ 0.5 = num ^ 0.5 and p ^ 0.5 <= q ^ 0.5.
// So we only need to check all odd numbers from 3 to num^0.5.
for (int i=3; i<=Math.sqrt(num); i+=2) {
if (num % i == 0) {
return false;
}
}

return true;
}
}