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?

Related problem: Reverse Integer

思路分析：这题比较简单，基本和Reverse Integer的思路类似，n不断右移位取末位，结果res不断左移位和n的末位取或，如此进行32次即可。考察简单的位运算。

AC Code

public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        //1145
        int res = n & 1;
        for(int i = 1; i < 32; i++){
            n = n>>1;
            res = res<<1;
            res = res | (n & 1);
        }
        return res;
    }
}