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?
32bit的一个逆置，很简单也不简单。通常想法是直接位操作，这样需要对称一下16次。我开始就是这样做的。
为什么说难呢？题目中说，如果需要多次调用该如何优化？
没想出来，看来discuss，有了下面这个方法，简介快速！
假设有数字abcdefgh，那么这个写法相当于做了这样的变换，每次对半交换。
abcdefgh -> efghabcd -> ghefcdab -> hgfedcba
class Solution { public: uint32_t reverseBits(uint32_t n) { n = (n >> 16) | (n << 16); n = ((n & 0xff00ff00) >> 8) | ((n & 0x00ff00ff) << 8); n = ((n & 0xf0f0f0f0) >> 4) | ((n & 0x0f0f0f0f) << 4); n = ((n & 0xcccccccc) >> 2) | ((n & 0x33333333) << 2); n = ((n & 0xaaaaaaaa) >> 1) | ((n & 0x55555555) << 1); return n; } };