Leetcode: Reverse Bits

题目：
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?

Related problem: Reverse Integer

思路分析：
从右到左取出每一位的数字，然后从左到右放置！注意位运算的神奇之处！

C++参考代码：

class Solution
{
public:
    uint32_t reverseBits(uint32_t n)
    {
        uint32_t result = 0;
        for (int i = 0; i < 32; ++i)
        {
            result <<= 1;//结果每次先左移一位，这样每次后面的数字 就能向前走一位
            if (n & 1) result |= 1;//如果n的末尾是1，则在result的后面修改为1
            //其实这里换成result ^= 1也是没问题的，因为0|1=1,0^1=1
            n >>= 1;//n右移，用于从右到左每次取后面的数字
        }
        return

因为C++整形数据所占的字节数会随着机器的不同而稍微有些区别，如果题目没有说明给定的无符号整形是4个字节，32位呢？我们可以通过数字1左移判断无符号整形的字节数。

C++参考代码：

class Solution
{
public:
    unsigned int reverseBits(unsigned int n)
    {
        unsigned int result = 0;
        //通过1左移判断无符号整形的字节数
        for (int i = 1; i != 0; i <<= 1)
        {
            result <<= 1;
            if (n & 1) result |= 1;
            n >>= 1;
        }
        return