# 190. Reverse Bits

## Description

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?

## Solutions

• public class Solution {
// you need treat n as an unsigned value
public int reverseBits(int n) {
int res = 0;
for (int i = 0; i < 32 && n != 0; ++i) {
res |= ((n & 1) << (31 - i));
n >>>= 1;
}
return res;
}
}

• class Solution {
public:
uint32_t reverseBits(uint32_t n) {
uint32_t res = 0;
for (int i = 0; i < 32; ++i) {
res |= ((n & 1) << (31 - i));
n >>= 1;
}
return res;
}
};

• class Solution:
def reverseBits(self, n: int) -> int:
res = 0
for i in range(32):
res |= (n & 1) << (31 - i)
n >>= 1
return res


• func reverseBits(num uint32) uint32 {
var ans uint32 = 0
for i := 0; i < 32; i++ {
ans |= (num & 1) << (31 - i)
num >>= 1
}
return ans
}

• /**
* @param {number} n - a positive integer
* @return {number} - a positive integer
*/
var reverseBits = function (n) {
let res = 0;
for (let i = 0; i < 32 && n > 0; ++i) {
res |= (n & 1) << (31 - i);
n >>>= 1;
}
return res >>> 0;
};


• impl Solution {
pub fn reverse_bits(mut x: u32) -> u32 {
let mut res = 0;
for _ in 0..32 {
res = (res << 1) | (x & 1);
x >>= 1;
}
res
}
}