經典問題8：c/c++ 程序設計 －－－bit位逆轉高效算法問題

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經典問題8：c/c++ 程序設計 －－－bit位逆轉高效算法問題
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        (3)面試題：寫一個函數，接受一個unsigned char的參數，返回一個unsigned char。
        函數要完成的功能是：把這個8bit的數從高到低翻轉過來。
        比如0x80轉換爲0x01, 0xA4轉換爲0x25。0xA4是 10100100 翻轉過來就是 00100101，也就是0x25。
解答：
     1    #include <iostream>
     2    #include <sys/time.h>
     3    using namespace std;
     4     
     5    //算法1，查表法，典型的空間換時間，在現代的CPU上，這種算法具有最快的速度。
     6    unsigned char reverse1(unsigned char c)
     7    {
     8            static unsigned char table[256] =
     9            {
        0x00,0x80,0x40,0xC0,0x20,0xA0,0x60,0xE0,0x10,0x90,0x50,0xD0,0x30,
        0xB0,0x70,0xF0,0x08,0x88,0x48,0xC8,0x28,0xA8,0x68,0xE8,0x18,0x98,
        0x58,0xD8,0x38,0xB8,0x78,0xF8,0x04,0x84,0x44,0xC4,0x24,0xA4,0x64,
        0xE4,0x14,0x94,0x54,0xD4,0x34,0xB4,0x74,0xF4,0x0C,0x8C,0x4C,0xCC,
        0x2C,0xAC,0x6C,0xEC,0x1C,0x9C,0x5C,0xDC,0x3C,0xBC,0x7C,0xFC,0x02,
        0x82,0x42,0xC2,0x22,0xA2,0x62,0xE2,0x12,0x92,0x52,0xD2,0x32,0xB2,
        0x72,0xF2,0x0A,0x8A,0x4A,0xCA,0x2A,0xAA,0x6A,0xEA,0x1A,0x9A,0x5A,
        0xDA,0x3A,0xBA,0x7A,0xFA,0x06,0x86,0x46,0xC6,0x26,0xA6,0x66,0xE6,
        0x16,0x96,0x56,0xD6,0x36,0xB6,0x76,0xF6,0x0E,0x8E,0x4E,0xCE,0x2E,
        0xAE,0x6E,0xEE,0x1E,0x9E,0x5E,0xDE,0x3E,0xBE,0x7E,0xFE,0x01,0x81,
        0x41,0xC1,0x21,0xA1,0x61,0xE1,0x11,0x91,0x51,0xD1,0x31,0xB1,0x71,
        0xF1,0x09,0x89,0x49,0xC9,0x29,0xA9,0x69,0xE9,0x19,0x99,0x59,0xD9,
        0x39,0xB9,0x79,0xF9,0x05,0x85,0x45,0xC5,0x25,0xA5,0x65,0xE5,0x15,
        0x95,0x55,0xD5,0x35,0xB5,0x75,0xF5,0x0D,0x8D,0x4D,0xCD,0x2D,0xAD,
        0x6D,0xED,0x1D,0x9D,0x5D,0xDD,0x3D,0xBD,0x7D,0xFD,0x03,0x83,0x43,
        0xC3,0x23,0xA3,0x63,0xE3,0x13,0x93,0x53,0xD3,0x33,0xB3,0x73,0xF3,               
        0x0B,0x8B,0x4B,0xCB,0x2B,0xAB,0x6B,0xEB,0x1B,0x9B,0x5B,0xDB,0x3B,
        0xBB,0x7B,0xFB,0x07,0x87,0x47,0xC7,0x27,0xA7,0x67,0xE7,0x17,0x97,
        0x57,0xD7,0x37,0xB7,0x77,0xF7,0x0F,0x8F,0x4F,0xCF,0x2F,0xAF,0x6F,
        0xEF,0x1F,0x9F,0x5F,0xDF,0x3F,0xBF,0x7F,0xFF               
    26            };
    27            return table[c];
    28    }
    29     
    30    //算法2，逆向移位，思路很簡單，代碼卻有點長.
    31    unsigned char reverse2( unsigned char c)
    32    {
    33            unsigned char r = 0;
    34     
    35            //r <<= 0,c >>= 0;
    36            r |= c&1;
    37     
    38            r <<= 1,c >>= 1;
    39            r |= c&1;
    40     
    41            r <<= 1,c >>= 1;
    42            r |= c&1;
    43     
    44            r <<= 1,c >>= 1;
    45     
    46            r |= c&1;
    47            r <<= 1,c >>= 1;
    48            r |= c&1;
    49     
    50            r <<= 1,c >>= 1;
    51            r |= c&1;
    52     
    53            r <<= 1,c >>= 1;
    54            r |= c&1;
    55     
    56            r <<= 1,c >>= 1;
    57            r |= c&1;
    58     
    59            return r;
    60    }
    61     
    62   
//算法3，逆向移位，和上一個算法同，但是用了循環，所以效率可能有點低。
    63    unsigned char reverse3( unsigned char c)
    64    {
    65            unsigned char r = 0;
    66          
    67            r |= c&1;
    68          
    69            for( int i = 0; i < 7; i++)
    70                    r <<= 1,c >>= 1,r |= c&1;     
    71     
    72            return r;
    73    }
    74     

//算法4，逐位判斷，看起來似乎比算法2更簡潔，但是因爲if語句牽涉到
//一個跳轉指令引起流水線重置的問題，在現在的CPU上不見得更快速。
    76    unsigned char reverse4( unsigned char c)
    77    {
    78            unsigned char r = 0;
    79     
    80            if( c&0x01 ) r |= 0x80;
    81            if( c&0x02 ) r |= 0x40;
    82            if( c&0x04 ) r |= 0x20;
    83            if( c&0x08 ) r |= 0x10;
    84            if( c&0x10 ) r |= 0x08;
    85            if( c&0x20 ) r |= 0x04;
    86            if( c&0x40 ) r |= 0x02;
    87            if( c&0x80 ) r |= 0x01;
    88          
    89            return r;
    90    }
    91     
//算法5，分段查表法。查表法雖然快，但是表有256個字節大，有些時候可//能顯得太大了。
//表太大，書寫不方便，而且看起來也比較凌亂。所以纔有了下面的算法，
//只用16字節的表。
    94    unsigned char reverse5( unsigned char c)
    95    {
    96            static unsigned char table[16] =
    97            {
    98     0x00,0x08,0x04,0x0C,0x02,0x0A,0x06,0x0E,0x01,0x09,0x05,0x0D,
0x03,0x0B,0x07,0x0F            
    99            };
   100            unsigned char r = 0;
   101     
   102            r |= (table[c&0xF]) << 4;
   103            r |= table[c>>4];     
   104          
   105            return r;
   106    }
   107     
   108    unsigned char reverse6( unsigned char c )
   109    {
   110            c = ( c & 0x55 ) << 1 | ( c & 0xAA ) >> 1;
   111            c = ( c & 0x33 ) << 2 | ( c & 0xCC ) >> 2;
   112            c = ( c & 0x0F ) << 4 | ( c & 0xF0 ) >> 4;
   113            return c;
   114    }
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性能結果：
$ ./a.out
msec[0]=477 （us:微妙）
msec[1]=4864
msec[2]=6078
msec[3]=1783
msec[4]=703
msec[5]=1003
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