Delphi判断一个字符串在另一个字符串中出现的次数 7个方法的效率对比。

unit Unit14;

interface

uses
  Winapi.Windows, Winapi.Messages, System.SysUtils, System.Variants, System.Classes, Vcl.Graphics,
  Vcl.Controls, Vcl.Forms, Vcl.Dialogs, System.StrUtils, Vcl.StdCtrls;

type
  TForm14 = class(TForm)
    Button1: TButton;
    procedure Button1Click(Sender: TObject);
  private
    { Private declarations }
  public
    { Public declarations }
  end;

var
  Form14: TForm14;

implementation

{$R *.dfm}

uses QString;

function GetStrNum(const SubStr, S: string): Integer;
var
  I: Integer;
begin
  Result := 0;
  I := PosEx(SubStr, S, 1);
  if I > 0 then
    Inc(Result)
  else
    Exit;

  while (I > 0) and (I < Length(S)) do
  begin
    I := PosEx(SubStr, S, I + 1);
    if I > 0 then
      Inc(Result);
  end;
end;

function StrPostime(Str:string;P:string):integer;
var i,n:integer;
begin
n:=0;
for i:=1 to length(str) do
begin
if copy(str,i,length(p))=P then n:=n+1;
end;
StrPostime:=n;
end;

function SubStrConut(mStr: string; mSub: string): Integer;
{ 返回子字符串出现的次数 }
begin
  Result :=
    (Length(mStr) - Length(StringReplace(mStr, mSub, '', [rfReplaceAll]))) div
    Length(mSub);
end; { SubStrConut }

function StrPosCount(subs:string;source:string):integer;
var
Str : string;
begin
Result := 0;
str := source;
while Pos(Subs,Str)<>0 do
begin
Delete(Str,Pos(Subs,Str),Length(Subs));
Inc(Result);
end;
end;


function InStrTimes(const AShortStr,ASourceStr: string): Integer;
var
  MyArrS: TArray<string>;
begin
  MyArrS := ASourceStr.Split([AShortStr], ExcludeEmpty);
  Exit(Length(MyArrS)-1);
end;

procedure TForm14.Button1Click(Sender: TObject);
var
  n: Cardinal;
  I: Integer;
begin
  //------------------------------------
  n := GetTickCount;
  for I := 0 to 1000000 do
  begin
    LeftStrCount('我靠你的我爱你的你不爱我的我不爱你的真心的很爱你的爱你的还需要理由吗','爱你的',false);
  end;
  n := GetTickCount - n;
  Text := IntToStr(n) + ' - ';


  //------------------------------------
  n := GetTickCount;
  for I := 0 to 1000000 do
  begin
    RightStrCount('我靠你的我爱你的你不爱我的我不爱你的真心的很爱你的爱你的还需要理由吗','爱你的',false);
  end;
  n := GetTickCount - n;
  Text := Text + IntToStr(n) + ' - ';


  //------------------------------------
  n := GetTickCount;
  for I := 0 to 1000000 do
  begin
    GetStrNum('爱你的','我靠你的我爱你的你不爱我的我不爱你的真心的很爱你的爱你的还需要理由吗');
  end;
  n := GetTickCount - n;
  Text := Text + IntToStr(n) + ' - ';


  //------------------------------------
  n := GetTickCount;
  for I := 0 to 1000000 do
  begin
    StrPostime('我靠你的我爱你的你不爱我的我不爱你的真心的很爱你的爱你的还需要理由吗','爱你的');
  end;
  n := GetTickCount - n;
  Text := Text + IntToStr(n) + ' - ';


  //------------------------------------
  n := GetTickCount;
  for I := 0 to 1000000 do
  begin
    SubStrConut('我靠你的我爱你的你不爱我的我不爱你的真心的很爱你的爱你的还需要理由吗','爱你的');
  end;
  n := GetTickCount - n;
  Text := Text + IntToStr(n) + ' - ';


  //------------------------------------
  n := GetTickCount;
  for I := 0 to 1000000 do
  begin
    StrPosCount('爱你的','我靠你的我爱你的你不爱我的我不爱你的真心的很爱你的爱你的还需要理由吗');
  end;
  n := GetTickCount - n;
  Text := Text + IntToStr(n) + ' - ';

  //------------------------------------
  n := GetTickCount;
  for I := 0 to 1000000 do
  begin
    InStrTimes('爱你的','我靠你的我爱你的你不爱我的我不爱你的真心的很爱你的爱你的还需要理由吗');
  end;
  n := GetTickCount - n;
  Text := Text + IntToStr(n);

end;

end.

 

 

可见第三个是最快的。

隔离出来第三个备份下：

function InStrTimes(const ASubStr,ASourceStr: string): Integer;
var
  I: Integer;
begin
  Result := 0;
  I := PosEx(ASubStr, ASourceStr, 1);
  if I > 0 then
  begin
    Inc(Result);
  end else begin
    Exit;
  end;
  while (I > 0) and (I < Length(ASourceStr)) do
  begin
    I := PosEx(ASubStr, ASourceStr, I + 1);
    if I > 0 then
    begin
      Inc(Result);
    end;
  end;
end;