# [Leet code] Reverse Bits

Leetcode: https://leetcode.com/problems/reverse-bits/

Problem:

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`

.

**Follow up**:

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

**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`

Solution:

`class Solution:`

# @param n, an integer

# @return an integer

def reverseBits(self, n):

test = list('{:032b}'.format(n))

test.reverse()

return int("".join(test),2)

Explanation:

‘{:032b}’.format(n) will convert an int to a binary number of 32 bits. We assign this to a list to make use of the reverse operator. Then we join this list into a string and int(number,2) will convert a binary back to a decimal.