CareerCup

public static void main(String[] args) {
// TODO Auto-generated method stub
for(int i = 2 ; i < 10000; i++) {
int j = i&(i-1);
if(j == 0) {
System.out.println("2s Power:: "+i);
}
}
}

public class Main {

/**
* @param args the command line arguments
*/
public static void main(String[] args) throws IOException {

BufferedReader reader=new BufferedReader(
new InputStreamReader(System.in) );



while(true)
{
System.out.print("input integer:");
String line= reader.readLine();

if(line.isEmpty())
{
break;
}

int digit=Integer.parseInt(line);
System.out.println("processing : "+ digit );

boolean isP=checkPowerOf2(digit);

if(isP)
{
System.out.println(digit+" is power of 2");
}else
{
System.out.println(digit+" is NOT power of 2");

}

}


reader.close();


}

private static boolean checkPowerOf2(int digit) {
long p2=2;

while(true)
{
if(p2>digit)
{
return true;
}


if(digit%p2!=0)
{
return false;
}

p2*=2;
}




}

}

count number of bits == 1.
if not==1 then not power of 2

Can't we check if(num%2 == 0) then print power of 2. Else not.

The person who posted the question has given answer in O(1) time complexity. I don't know, why HertzLee answered it differently.

check for 0 must be there. ) is not power of 2.
if((!num) && (!(num & num-1))

For all the folks who think the answer posted with the question is right: IT IS PLAIN WRONG.
As Tulley pointed out it should be "if((num) && (!(num & num-1))"
But looking at blueskin's solution: counting number of bits and if equals to 1 or break counting when greater than 1 is the best solution because counting bits is faster than applying bit operations between two integers

If is allowed to use library functions : Find out the log(base 2 ) of the input, check whether integer or not.