[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.