fast array search with delphi

[Bjoerk]

Delphi-Quellcode:

function equalpoints(p1x, p1y, p2x, p2y: integer): boolean;
begin
  Result:= ((p1x = p2x) and (p1y = p2y));
end;


function IndexOfVector(const sx, sy, ex, ey: integer): integer; 
var
  M: integer; 
begin
  Result:= 0; 
  for M:= 1 to VNum-1 do
    if equalpoints (sx, sy, V[M].sx, V[M].sy) then
      if equalpoints (ex, ey, V[M].ex, V[M].ey) then
      begin
        Result:= M; 
        Break; 
      end; 
end; 


procedure removevector(mm, mm2: integer); 
var
  P, N: integer; 
begin
  P:= V[mm].prev; 
  V[P].next:= V[mm2].next; 

  N:= V[mm2].next; 
  V[N].prev:= P; 
end; 


procedure removevectors(M, M2: integer);
begin
  removevector(M, M2); 
  removevector(M2, M); 

  V[M].status:= -1; 
  V[M2].status:= -1; 
end; 


procedure addsquarevector(J, K: integer);
var
  M: integer; 
begin
  Vnum:= Vnum + 1; 
  V[Vnum].prev:= Vnum + 3; 
  V[Vnum].sx:= J; 
  V[Vnum].sy:= K; 
  V[Vnum].ex:= J + 1; 
  V[Vnum].ey:= K; 
  V[Vnum].next:= Vnum + 1; 
  V[Vnum].status:= 0; 
  M:= IndexOfVector (V[Vnum].sx, V[Vnum].sy, V[Vnum].ex, V[Vnum].ey); 
  if M <> 0 then
  begin
    removevectors(M, Vnum); 
    Vnum:= Vnum - 1
  end; 

  Vnum:= Vnum + 1; 
  V[Vnum].prev:= Vnum - 1; 
  V[Vnum].sx:= J + 1; 
  V[Vnum].sy:= K; 
  V[Vnum].ex:= J + 1; 
  V[Vnum].ey:= K + 1; 
  V[Vnum].next:= Vnum + 1; 
  V[Vnum].status:= 0; 
  M:= IndexOfVector (V[Vnum].sx, V[Vnum].sy, V[Vnum].ex, V[Vnum].ey); 
  if M <> 0 then
  begin
    removevectors(M, Vnum); 
    Vnum:= Vnum - 1
  end; 

  Vnum:= Vnum + 1; 
  V[Vnum].prev:= Vnum - 1; 
  V[Vnum].sx:= J + 1; 
  V[Vnum].sy:= K + 1; 
  V[Vnum].ex:= J; 
  V[Vnum].ey:= K + 1; 
  V[Vnum].next:= Vnum + 1; 
  V[Vnum].status:= 0;
  M:= IndexOfVector (V[Vnum].sx, V[Vnum].sy, V[Vnum].ex, V[Vnum].ey);
  if M <> 0 then
  begin
    removevectors(M, Vnum);
    Vnum:= Vnum - 1
  end;

  Vnum:= Vnum + 1;
  V[Vnum].prev:= Vnum - 1;
  V[Vnum].sx:= J;
  V[Vnum].sy:= K + 1;
  V[Vnum].ex:= J;
  V[Vnum].ey:= K;
  V[Vnum].next:= Vnum - 3;
  V[Vnum].status:= 0;
  M:= IndexOfVector (V[Vnum].sx, V[Vnum].sy, V[Vnum].ex, V[Vnum].ey);
  if M <> 0 then
  begin
    removevectors(M, Vnum);
    Vnum:= Vnum - 1
  end;
end;

Ob so aber tatsächlich eine Zeitersparnis entsteht, müßte man erst ausprobieren.