190. Reverse Bits #

Problem Statement #

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: #

impl Solution {
    pub fn reverse_bits(x: u32) -> u32 {
        let (mut x, mut res) = (x, 0);

        // iterate 32 times, once for each bit
        for _ in 0..32 {
            // left-shift res by 1 bit to make room for the next bit
            res <<= 1;
            // adding the least significant bit of x to the result
            // using bitwise AND with 1
            res += x & 1;
            // right-shift x by 1 bit, discarding the least significant bit
            // that was just added to the result
            x >>= 1;
        }

        res
    }
}

... #