leetcode- Power of Two

#. Problem

Power of Two

1. My approach

brute force

Even if I divided it by 2, the whole process will only take 32 times even on the worst case.

bool isPowerOfTwo(int n) {
    if(n <= 0) return false;

    while(n%2 == 0){
        n /= 2;
    }

    if(n == 1) return true;
    else return false;
}

2. Solution

cleaver

Very easy and efficient.

Based on

1. Binary code of power of 2 would only have one 1. ex) 000100
2. power of 2 can’t be smaller than 0

You can come up with the solution below.

bool isPowerOfTwo(int n) {
   if(n<=0) return false;
   return (n & (n-1))? false:true;
}

0 0 0 1 0 0 0 0 0
0 0 0 0 1 1 1 1 1
&
0 0 0 0 0 0 0 0 0

n & (n - 1) cannot be true when n is power of 2.