# 190.reverse-bits

## Statement

• Difficulty: Easy
• Tag: `位运算` `分治`

• 请注意，在某些语言（如 Java）中，没有无符号整数类型。在这种情况下，输入和输出都将被指定为有符号整数类型，并且不应影响您的实现，因为无论整数是有符号的还是无符号的，其内部的二进制表示形式都是相同的。
• 在 Java 中，编译器使用二进制补码记法来表示有符号整数。因此，在 示例 2 中，输入表示有符号整数 `-3`，输出表示有符号整数 `-1073741825`

``````输入：n = 00000010100101000001111010011100

因此返回 964176192，其二进制表示形式为 00111001011110000010100101000000。``````

``````输入：n = 11111111111111111111111111111101

因此返回 3221225471 其二进制表示形式为 10111111111111111111111111111111 。``````

• 输入是一个长度为 `32` 的二进制字符串

• Difficulty: Easy
• Tag: `Bit Manipulation` `Divide and Conquer`

Reverse bits of a given 32 bits unsigned integer.

Note:

• Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's internal binary representation is the same, whether it is signed or unsigned.
• In Java, the compiler represents the signed integers using 2's complement notation. Therefore, in Example 2 above, the input represents the signed integer `-3` and the output represents the signed integer `-1073741825`.

Example 1:

``````Input: n = 00000010100101000001111010011100
Output:    964176192 (00111001011110000010100101000000)
Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.
``````

Example 2:

``````Input: n = 11111111111111111111111111111101
Output:   3221225471 (10111111111111111111111111111111)
Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10111111111111111111111111111111.
``````

Constraints:

• The input must be a binary string of length `32`

Follow up: If this function is called many times, how would you optimize it?

## Solution

``````class Solution:
def reverseBits(self, n: int) -> int:
return int('{:032b}'.format(n)[::-1], 2)
``````