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.

Constraints:

• The input must be a binary string of length 32

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**.

Solve¶

Here is what i done:

• Turn number n to binary string using bin(n)
• Padding 0 to get enough 32 bit, by adding (1<<32) and remove first bit of the string bin(n) (And 0b string from the result, which in total is 3 character)
• Revert the product string using [::-1]
• And finally turn the revert binary string back to number using int(<binary_string>, base=2)
  

  class Solution:
      def reverseBits(self, n: int) -> int:
          return int`-1], base=2`
  

Or just simple adding it bit by bit, using binary manipulation

uint32_t reverseBits(uint32_t n) {
    uint32_t r_num = 0;
    for (int i = 0; i < 32; i++) {
        r_num <<= 1; //r_num = r_num * 2
        r_num += (n >> i) & 1; // r_num = r_num + n[32-i] 
    }
    return r_num;
}