Reverse Bits
題意：
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).
Follow up:
If this function is called many times, how would you optimize it?
解題思路：
利用一個count值從31往1到數，來決定取出來的bit要往右shift幾格，程式碼如下：
public class Solution {
// you need treat n as an unsigned value
public int reverseBits(int n) {
int sum = 0;
int count = 31;
while (n != 0) {
if ((n & 1) == 1) {
sum += 1 << count;
}
count--;
n >>>= 1;
}
return sum;
}
}