h*****g
发帖数: 944 | 1 given 10001110
want to get 01110001
只能用 and/or这两种 bit operation | g*******s
发帖数: 490 | 2 用& | 和 bit shift的方法有，只用& |的还真不知道
unsigned __int32 reversing_bits(unsigned __int32 input)
{
// complixity O(log [no.of.bits]) = O(1)
// On 32 bit machines it takes 5 steps (logical)
// Step 1
// Mask bit positions 0, 2, 4... shift LEFT this masked number by one
// Mask bit positions 1, 3, 5... shift RIGHT this masked number by one
input = (input & 0x55555555) << 1 | (input & 0xAAAAAAAA) >> 1;
// Step 2
// Mask bit positions 01, 45, 89... shift LEFT this masked number by two
// Mask bit positions 23, 67... shift RIGHT this masked number by two
input = (input & 0x33333333) << 2 | (input & 0xCCCCCCCC) >> 2;
// Step 3
input = (input & 0x0F0F0F0F) << 4 | (input & 0xF0F0F0F0) >> 4;
// Step 4
input = (input & 0x00FF00FF) << 8 | (input & 0xFF00FF00) >> 8;
// Sept 5
input = (input & 0x0000FFFF) << 16 | (input & 0xFFFF0000) >> 16;
return input;
} | g*********s
发帖数: 1782 | 3 没空间开销约束的话好办啊。
就一位一位的&，记录下每一位，再倒过来一位一位or即可。
【在 h*****g 的大作中提到】
: given 10001110
: want to get 01110001
: 只能用 and/or这两种 bit operation
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