《Delphi 算法与数据结构》学习与感悟[3]: 获取一个字节中非空位的个数

一个字节有 8 个位, 这些位可能是 0 也可能是 1; 现在要算出一个字节中是 1 的位共有多少个.

第一种方法是一个函数;
第二种方法笨了点, 是先把 256 种可能值给一个数组, 随时调取.

第一种方法虽然灵巧, 但不如第二种方法快(作者书中说: 在非特殊情况下, 一般要快到 10 倍左右);
第二种方法虽然快捷, 并且使用方便, 但要以 256 个字节的数组空间为代价.

---

unit Unit1;



interface



uses

  Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls, Forms,

  Dialogs, StdCtrls;



type

  TForm1 = class(TForm)

    Button1: TButton;

    procedure Button1Click(Sender: TObject);

  end;



var

  Form1: TForm1;



implementation



{$R *.dfm}



{方法1: 获取函数}

function GetByteBits(x: Byte): Byte;

begin

  Result := 0;

  while x <> 0 do

  begin

    if Odd(x) then Inc(Result);

    x := x shr 1;

  end;

end;



{方法2: 把所有可能的值放在一个常数数组}

const

  BitArr: array[0..MAXBYTE] of Byte = (

    0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,

    1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,

    1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,

    2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,

    1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,

    2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,

    2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,

    3,4,4,5,4,5,5,6,4,5,5,6,5,6,6,7,4,5,5,6,5,6,6,7,5,6,6,7,6,7,7,8);



{测试}

procedure TForm1.Button1Click(Sender: TObject);

var

  b,num: Byte;

begin

  b := 255;

  num := GetByteBits(b);      {使用函数获取}

  ShowMessage(IntToStr(num)); {8}

  num := BitArr[b];           {直接使用数组获取}

  ShowMessage(IntToStr(num)); {8}



  b := 254;

  num := GetByteBits(b);      {使用函数获取}

  ShowMessage(IntToStr(num)); {7}

  num := BitArr[b];           {直接使用数组获取}

  ShowMessage(IntToStr(num)); {7}

end;



end.


 
   

那个小函数, 琢磨了半天才明白(惭愧); 以后判断其他数也没问题了, 譬如判断 Integer:

---

function GetIntBits(x: Integer): Byte;

begin

  Result := 0;

  while x <> 0 do

  begin

    if Odd(x) then Inc(Result);

    x := x shr 1;

  end;

end;