###### tags: `Leetcode` `easy` `math` `bit` `python` `c++`

# 231. Power of Two

## [題目連結:] https://leetcode.com/problems/power-of-two/description/

## 題目:
Given an integer ```n```, return ```true``` if it is a power of two. Otherwise, return ```false```.

An integer ```n``` is a power of two, if there exists an integer ```x``` such that ```n == 2x```.

**Example 1:**

Input: n = 1
Output: true
Explanation: 20 = 1

**Example 2:**

Input: n = 16
Output: true
Explanation: 24 = 16

**Example 3:**

Input: n = 3
Output: false

**Follow up:** Could you solve it without loops/recursion?

## 解題想法:
* 此題為判斷給的數是否為2的冪次方
* 簡易解:
* 直接while迴圈累除2判斷即可
* 解Follow up:
* 使用**Bit**判斷
* 若n為2的冪次方:
* 其二進制為:10...0
* 則(n-1)為二進制為:01111....1
* **兩者做and為0**

## Python:

class Solution(object):
    def isPowerOfTwo(self, n):
        """
        :type n: int
        :rtype: bool
        """
        #法1 
        #若n為2的冪次方，二進制為 10...0，則(n-1)為01111....1 兩者做and為0
        if n<=0:
            return False
        else:
            # 0為fasle 所以要加個not才負負的正
            return not (n&(n-1))
        '''
        #法2
        while n:
            if n%2 ==0:
                n = n//2
            elif n==1:
                return True
            else:
                return False
        '''

if __name__ == '__main__':
    n = 16
    result = Solution()
    ans = result.isPowerOfTwo(n)
    print(ans)

## C++:

class Solution {
public:
    bool isPowerOfTwo(int n) {
        if (n<=0)
            return false;
        else
            return !(n&(n-1));
    }
};