190. 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. 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

190. Reverse Bits
struct Solution;

impl Solution {
    fn reverse_bits(x: u32) -> u32 {
        let mut res = 0;
        for i in 0..32 {
            let bit = x & 1 << i;
            let travel = (31 - i as i32) - i as i32;
            if travel > 0 {
                res |= bit << travel;
            } else {
                res |= bit >> -travel;
            }
        }
        res
    }
}

#[test]
fn test() {
    let n = 0b00000010100101000001111010011100;
    let res = 964176192;
    assert_eq!(Solution::reverse_bits(n), res);
    let n = 0b11111111111111111111111111111101;
    let res = 3221225471;
    assert_eq!(Solution::reverse_bits(n), res);
}