简单问题，50分给一人，留下的其它人分(先给出方法的50分)

[masterjames]

有一字符串 如:STR1:='1.02 aaa bb c ddd bm' 中间用空格分开,但空格的数目不相同,
如,1.02和 AAA 中间可能有3个,而BB与AAA中只有一个. 求一简单的算法,将以上
数据分开逐个添加到TStringList的变量list 中
50分给一人，留下的其它人分(先给出方法的50分),通过者立马给分.

[JeffChung]

delimitedText

[JeffChung]

var
  str1:string;
  list:TStringList;
  i:integer;
begin
  str1:='1.02  aaa bb    c    ddd bm' ;
  list := TStringList.Create;
  list.DelimitedText := str1;
  memo1.Lines.AddStrings(list);
  list.Free;
end;

[liangpei2008]

var
  s: String;
  List: TStringList;
begin
  s := '1.02  aaa bb    c    ddd bm' ;
  List := TStringList.Create;
  ExtractStrings([' '],[' '],PChar(s),List);
  Showmessage(List[4]);
  List.Free;
end;

[iamduo]

http://blog.sina.com.cn/s/blog_40ac71670100anjg.html

[liuhengwinner]

用状态机，详见 Delphi 算法与数据结构一书！我写过的，现在手边没代码，明天粘给你！

[liangpei2008]

Uses
  IdStrings;
var
  s: String;
  List: TStringList;
begin
  s := '1.02  aaa bb    c    ddd bm' ;
  List := TStringList.Create;
  SplitColumns(s,List,' ');
  List.Free;
end;

[liuhengwinner]

这么做的话还得去除空格！有的间隔是两个或三个空格！不如状态机灵活！

[liuhengwinner]

Function TCountWord.ExtractWords(const S: string): TStringDynArray;
type
TStates = (ScanNormal,ScanQuoted,ScanPunctuation);
const
WordDelim = ' !<>[](){},./?;~:-+=*@#$%^&"12345567890';
var
State: TStates;
Inx: integer;
Ch: Char;
CurWord: string;

procedure AddToList(Const aWord: string);
var
aLen: integer;
begin
aLen := Length(Result);
SetLength(Result,aLen+1);
Result[aLen] := lowerCase(aWord);
end;
begin
SetLength(Result,0);
State := ScanNormal;
CurWord :='';
//读入串中所有字符
for Inx := 1 to Length(S) do
begin
//获取下一个字符
Ch := S[Inx];
//对于状分情况处理
case State of
ScanNormal:
begin
if Pos(Ch,WordDelim)<>0 then
begin
if (CurWord<>'') then
begin
AddToList(CurWord);
CurWord := '';
end;
state := ScanPunctuation;
end else
begin
CurWord := CurWord+Ch;
end;
end;
ScanQuoted:
begin
CurWord := CurWord+Ch;
State := ScanNormal;
end;
ScanPunctuation:
begin
if Pos(Ch,WordDelim) = 0 then
begin
CurWord := Ch;
State := ScanNormal;
end;
end;

end;
end;
if CurWord<>'' then
AddToList(CurWord);
end;