How to wit to determine whether an integer is 2 n-th power

The beauty of programming-Microsoft technical Interview Experience the book details several methods that are of interest to see.

I'm only writing two methods that are easier to understand.

First Kind

2 10

4 100

8 1000

。。。。。

Found that the rule is not, 2 of the N-second side into the binary after the first 1, the rest of the bits are 0.

So, we can do this with the flattering, assuming the input is int:

Boolean powerof2 (int n)
{return
      ((n& (n-1)) ==0)
}

The second is to take this number to 2 modulo 0, and then divide 2, and then 2 modulo, until the number is finally 2, if you can't do this is not 2 of the N-Power. This method if not understand can take 56 of this number, to follow the process of their own calculation. We can understand the secret of this.

int count = 1;
while (i)
{
    if (2 = i)        //Note that the final Judgment of 2 is the key
    {
        System.out.println ("yes");  
        break;
    if (0 = = i%2)
    {
       i = I/2;
       Count + +;
     }
     else
     {
        System.out.print ("no");
        break;
      }
}

There are a lot of ways, we can go to divergent thinking.