# slt

首先明确1点，负数和零不可能是 power of two

## Solution1 不断 \* 2 比较

* 时间复杂度 O(logn)
* 空间复杂度 O(1)

## Solution2 bit

* 时间复杂度 O(1)
* 空间复杂度 O(1)

n为2的幂次方，则恒有：

* n 二进制最高位为 1，其余所有位为 0
* n−1 二进制最高位为 0，其余所有位为 1

-> `n & (n-1) == 0`

| 2^x | n    | n-1  | n&(n-1) |
| --- | ---- | ---- | ------- |
| 2^0 | 1    | 0    | 0       |
| 2^1 | 10   | 01   | 0       |
| 2^2 | 100  | 011  | 0       |
| 2^3 | 1000 | 0111 | 0       |

但是注意思路2跑的时间实际上大于思路1哦


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