Most Elegant Way to write isPrime in java
- by Anantha Kumaran
public class Prime {
public static boolean isPrime1(int n) {
if (n <= 1) {
return false;
}
if (n == 2) {
return true;
}
for (int i = 2; i <= Math.sqrt(n) + 1; i++) {
if (n % i == 0) {
return false;
}
}
return true;
}
public static boolean isPrime2(int n) {
if (n <= 1) {
return false;
}
if (n == 2) {
return true;
}
if (n % 2 == 0) {
return false;
}
for (int i = 3; i <= Math.sqrt(n) + 1; i = i + 2) {
if (n % i == 0) {
return false;
}
}
return true;
}
}
public class PrimeTest {
public PrimeTest() {
}
@Test
public void testIsPrime() throws IllegalArgumentException, IllegalAccessException, InvocationTargetException {
Prime prime = new Prime();
TreeMap<Long, String> methodMap = new TreeMap<Long, String>();
for (Method method : Prime.class.getDeclaredMethods()) {
long startTime = System.currentTimeMillis();
int primeCount = 0;
for (int i = 0; i < 1000000; i++) {
if ((Boolean) method.invoke(prime, i)) {
primeCount++;
}
}
long endTime = System.currentTimeMillis();
Assert.assertEquals(method.getName() + " failed ", 78498, primeCount);
methodMap.put(endTime - startTime, method.getName());
}
for (Entry<Long, String> entry : methodMap.entrySet()) {
System.out.println(entry.getValue() + " " + entry.getKey() + " Milli seconds ");
}
}
}
I am trying to find the fastest way to check whether the given number is prime or not. This
is what is finally came up with. Is there any better way than the second implementation(isPrime2).
