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# 231.Power of Two

**231.Power of Two**

难度:Easy

> 给定一个整数，编写一个函数来判断它是否是 2 的幂次方。

```
示例 1:

输入: 1
输出: true
解释: 20 = 1
示例 2:

输入: 16
输出: true
解释: 24 = 16
示例 3:

输入: 218
输出: false
```

直接位运算，二的整数次幂，必然只有一个位上是1，其他位为零。

```
class Solution {
public:
    bool isPowerOfTwo(int n) {
        if(n==0) return false;
        while((n&1) ==0)
        {
            n>>=1;
        }
     //   cout<<n<<endl;
        return n==1;
    }
};
```

> 执行用时 :8 ms, 在所有 C++ 提交中击败了60.55%的用户\
> 内存消耗 :8.1 MB, 在所有 C++ 提交中击败了27.77%的用户


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