# LeetCode – Reverse Bits (Java)

Problem

Reverse bits of a given 32 bits unsigned integer.

For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000).

If this function is called many times, how would you optimize it?

Related problem: Reverse Integer

Java Solution

```public int reverseBits(int n) { for (int i = 0; i < 16; i++) { n = swapBits(n, i, 32 - i - 1); }   return n; }   public int swapBits(int n, int i, int j) { int a = (n >> i) & 1; int b = (n >> j) & 1;   if ((a ^ b) != 0) { return n ^= (1 << i) | (1 << j); }   return n; }```
Category >> Algorithms >> Interview >> Java
If you want someone to read your code, please put the code inside <pre><code> and </code></pre> tags. For example:
```<pre><code>
String foo = "bar";
</code></pre>
```
• Alik Elzin

I think this is more easier to understand.
Pull from the lower end and push to the reversed string.

``` int reverseBits(int num) { int reverse = 0;```

``` ```

``` for (int i = 0 ; i >= 1) { reverse <<= 1; if (num % 2 == 1) reverse += 1; } } ```

• Rashmith Koundinya

what is x ?

• Zhongyuan Zhang

is this the optimised solution?

map = {0: 0, 1: 8, 2: 4, 3: 12, 4: 2, 5: 10, 6: 6, 7: 14, 8: 1, 9: 9, 10: 5, 11: 13, 12: 3, 13: 11, 14: 7, 15: 15}
res = 0
for i in range(8):
res <>= 4
return res

• Walden

public class Solution {
// you need treat n as an unsigned value
public int reverseBits(int n) {