reverse bit （！！！）

如果只是reverse一個integer 可以由兩種方法

1. 兩個bit交換

typedef unsigned int uint;
uint swapBits(uint x, uint i, uint j) {
  uint lo = ((x >> i) & 1);
  uint hi = ((x >> j) & 1);
  if (lo ^ hi) {
    x ^= ((1U << i) | (1U << j));
  }
  return x;
}
 
uint reverseXor(uint x) {
  uint n = sizeof(x) * 8;
  for (uint i = 0; i < n/2; i++) {
    x = swapBits(x, i, n-i-1);
  }
  return x;
}

注意要寫成 1U

2. 二分法

 Let us use an example of n == 8 (one byte) to see how this works:

      01101001
    /         \
   0110      1001
  /   \     /   \
 01   10   10   01
 /\   /\   /\   /\
0 1  1 0  1 0  0 1

The first step is to swap all odd and even bits. After that swap consecutive pairs of bits, and so on…
Therefore, only a total of log(n) operations are necessary.

uint reverseMask(uint x) {
  assert(sizeof(x) == 4); // special case: only works for 4 bytes (32 bits).
  x = ((x & 0x55555555) << 1) | ((x & 0xAAAAAAAA) >> 1);
  x = ((x & 0x33333333) << 2) | ((x & 0xCCCCCCCC) >> 2);
  x = ((x & 0x0F0F0F0F) << 4) | ((x & 0xF0F0F0F0) >> 4);
  x = ((x & 0x00FF00FF) << 8) | ((x & 0xFF00FF00) >> 8);
  x = ((x & 0x0000FFFF) << 16) | ((x & 0xFFFF0000) >> 16);
  return x;
}

如果要reverse很多integer，可以用上述方法再加hash查表，表的大小爲2^n， n爲unsigned int的bit 長度