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`.

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);
}
``````