# 190 Java: Reverse Bits – Easy

### 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 as00111001011110000010100101000000).

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

### Thoughts:

There could be different ways to solve this problem.

The solution below is using a helper function swapBits which makes things a littler easier.

swapBits is responsible for swapping bits by given bit index.

### Solutions:

``````public class Solution {
// you need treat n as an unsigned value
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;
}
}
``````
``````public class Solution {
// you need treat n as an unsigned value
public int reverseBits(int n) {
int result = 0;
for (int i = 0; i < 32; i ++) {
int digit = n & mask;
result = (result << 1);
if (digit != 0) {
result = result + 1;
}