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 = 00000010100101000001111010011100Output:964176192 (00111001011110000010100101000000)Explanation:The input binary string00000010100101000001111010011100represents the unsigned integer 43261596, so return 964176192 which its binary representation is00111001011110000010100101000000.

**Example 2:**

Input:n = 11111111111111111111111111111101Output:3221225471 (10111111111111111111111111111111)Explanation:The input binary string11111111111111111111111111111101represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is10111111111111111111111111111111.

**Constraints:**

- The input must be a
**binary string**of length`32`

```
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);
}
```