## Question

http://www.codewars.com/kata/534d0a229345375d520006a0/train/javascript

Write a function that determines if given number is a power of two. A power of two means a number of the form 2^n where n is an integer, i.e. the result of exponentiation with number two as the base and integer n as the exponent. I.e. 1024 is a power of two: it 2^10.

isPowerOfTwo(4096) // -> true isPowerOfTwo(333) // -> false

Pay attention: hidden tests are using extremmely big numbers

## My Solution

function isPowerOfTwo(n){ if(n==2){return true;} else if(n%2==1){return false;} else {return isPowerOfTwo(n/2);} }

처음엔 재귀함수를 돌 때 앞에

`return`

을 쓰지 않으니

마지막에 불리우는`return`

문에서 함수를 빠져나가는게 아니라 재귀함수를 빠져나가버렸다.

그래서 계속`undefined`

가 나왔다.

## @BinaryPanda ‘s Solution

function isPowerOfTwo(n) { return n && !(n & (n - 1)); }

이진법 연산이다. 이런 사람들 보면 내가 미워진다.

For example: 1 is 000001 2 is 000010 4 is 000100 8 is 001000 16 is 010000 32 is 100000

So, we need a way to verify that only a single digit is set. A bitwise and (the single &) will compare the digits in the positions they are in. So 10 & 01 is 00, but 11 & 10 is 10. Only if both spots contain a 1, does the result contain a 1.

If a number is a power of two, subtracting one will result in something like this:

8 = 001000 7 = 000111

7 & 8 will be 0, because they share no 1’s in the same spot. Then by using a !, we invert the answer.

What if the number is not a power of two? Like what about 7?

7 = 000111 6 = 000110 7 & 6 = 000110 (6)

## @PandaCoder’s Solution

function isPowerOfTwo(n){ return Math.log(n)/Math.log(2) % 1==0; }