## Algorithm

Problem Name: 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`.

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`

## Code Examples

### #1 Code Example with C Programming

```Code - C Programming```

``````
uint32_t reverseBits(uint32_t n) {
/*
uint32_t k, i;
k = 0;
for (i = 0; i < 32; i ++) {
k = (k << 1) | (n & 1);
n = n >> 1;
}
return k;
*/
n = ((n & 0xffff0000) >> 16) | ((n & 0x0000ffff) << 16);
n = ((n & 0xff00ff00) >> 8)  | ((n & 0x00ff00ff) << 8);
n = ((n & 0xf0f0f0f0) >> 4)  | ((n & 0x0f0f0f0f) << 4);
n = ((n & 0xcccccccc) >> 2)  | ((n & 0x33333333) << 2);
n = ((n & 0xaaaaaaaa) >> 1)  | ((n & 0x55555555) << 1);

return n;
}
``````
Copy The Code &

Input

cmd
n = 00000010100101000001111010011100

Output

cmd
964176192 (00111001011110000010100101000000)

### #2 Code Example with Java Programming

```Code - Java Programming```

``````
public class Solution {
// you need treat n as an unsigned value
public int reverseBits(int n) {
if (n == 0) {
return 0;
}
int result = 0;
for (int i = 0; i < 32; i++) {
result <<= 1;
if ((n & 1) == 1) {
result++;
}
n >>= 1;
}
return result;
}
}
``````
Copy The Code &

Input

cmd
n = 00000010100101000001111010011100

Output

cmd
964176192 (00111001011110000010100101000000)

### #3 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const reverseBits = function(n) {
let r = 0;
for (let i = 0; i < 32; i++) {
if (n & 1) {
r = r | 1;
}
if (i !== 31) {
r = r << 1;
n = n >> 1;
}
}
return r >>> 0;
};
``````
Copy The Code &

Input

cmd
n = 11111111111111111111111111111101

Output

cmd
3221225471 (10111111111111111111111111111111)

### #4 Code Example with C# Programming

```Code - C# Programming```

``````
namespace LeetCode
{
public class _0190_ReverseBits
{
public uint reverseBits(uint n)
{
n = (n >> 16) | (n << 16);
n = ((n & 0xff00ff00) >> 8) | ((n & 0x00ff00ff) << 8);
n = ((n & 0xf0f0f0f0) >> 4) | ((n & 0x0f0f0f0f) << 4);
n = ((n & 0xcccccccc) >> 2) | ((n & 0x33333333) << 2);
n = ((n & 0xaaaaaaaa) >> 1) | ((n & 0x55555555) << 1);
return n;
}
}
}
``````
Copy The Code &

Input

cmd
n = 11111111111111111111111111111101

Output

cmd
3221225471 (10111111111111111111111111111111)