190 Reverse Bits

Reverse Bits

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](https://en.wikipedia.org/wiki/Two%27s_complement). 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?

class Solution:
    def reverseBits(self, n: int) -> int:
        first_bit = 1 #2147483648 # unsigned 2 pow 31
        n_backup = n
        result = 0
        for k in range(32):
            temp = first_bit & n
            if temp == first_bit: # was 1
                result = result << 1
                result = result | 1
            else : # was 0
                result = result << 1
            print(k,result, n)
            n = n >> 1
        return result