Reverse bits of a given 32 bits unsigned integer.
Note:
-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:
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);
}