我想做的是确定括号的顺序是否正确。例如([][[]]<<>>)
有效,但是][]<<(>>)
不是。
我有一个工作版本,但它的效率很糟糕,当它得到 1000+ 括号时,它的速度简直太慢了。我希望有人能提出一些可能的改进或其他方法。
这是我的代码:
program Codex;
const
C_FNAME = 'zavorky.in';
var TmpChar : char;
leftBrackets, rightBrackets : string;
bracketPos : integer;
i,i2,i3 : integer;
Arr, empty : array [0..10000] of String[2];
tfIn : Text;
result : boolean;
begin
leftBrackets := ' ( [ /* ($ <! << ';
rightBrackets := ' ) ] */ $) !> >> ';
i := 0;
result := true;
Assign(tfIn, C_FNAME);
Reset(tfIn);
{ load data into array }
while not eof(tfIn) do
begin
while not eoln(tfIn) do
begin
read(tfIn, TmpChar);
if (TmpChar <> ' ') then begin
if (TmpChar <> '') then begin
Arr[i] := Arr[i] + TmpChar;
end
end
else
begin
i := i + 1;
end
end;
i2 := -1;
while (i2 < 10000) do begin
i2 := i2 + 1;
{if (i2 = 0) then
writeln('STARTED LOOP!');}
if (Arr[i2] <> '') then begin
bracketPos := Pos(' ' + Arr[i2] + ' ',rightBrackets);
if (bracketPos > 0) then begin
if (i2 > 0) then begin
if(bracketPos = Pos(' ' + Arr[i2-1] + ' ',leftBrackets)) then begin
{write(Arr[i2-1] + ' and ' + Arr[i2] + ' - MATCH ');}
Arr[i2-1] := '';
Arr[i2] := '';
{ reindex our array }
for i3 := i2 to 10000 - 2 do begin
Arr[i3 - 1] := Arr[i3+1];
end;
i2 := -1;
end;
end;
end;
end;
end;
{writeln('RESULT: ');}
For i2:=0 to 10 do begin
if (Arr[i2] <> '') then begin
{write(Arr[i2]);}
result := false;
end;
{else
write('M');}
end;
if (result = true) then begin
writeln('true');
end
else begin
writeln('false');
end;
result := true;
{ move to next row in file }
Arr := empty;
i := 0;
readln(tfIn);
end;
Close(tfIn);
readln;
end.
文件 zavorky.in 中的输入数据如下所示:
<< $) >> << >> ($ $) [ ] <! ( ) !>
( ) /* << /* [ ] */ >> <! !> */
我确定每一行是否有效。一行中括号的最大数量为 10000。
最佳答案
您从文件中读取了字符。以字节方式读取文件非常慢。您需要优化读取字符串(缓冲区)的方式,或者首先将文件加载到内存中。
下面我提出另一种方法来处理获取的字符串。
首先,我声明常量来声明您可能拥有的括号:
const
OBr: array [1 .. 5{6}] of string = ('(', '[', '/*', '<!', '<<'{, 'begin'});
CBr: array [11 .. 15{16}] of string = (')', ']', '*/', '!>', '>>'{, 'end'});
我决定这样做,因为现在您不再受限于括号表达式的长度和/或括号类型的数量。每个右括号和相应的左括号的指数差等于 10。
这是该函数的代码:
function ExpressionIsValid(const InputStr: string): boolean;
var
BracketsArray: array of byte;
i, Offset, CurrPos: word;
Stack: array of byte;
begin
result := false;
Setlength(BracketsArray, Length(InputStr) + 1);
for i := 0 to High(BracketsArray) do
BracketsArray[i] := 0; // initialize the pos array
for i := Low(OBr) to High(OBr) do
begin
Offset := 1;
Repeat
CurrPos := Pos(OBr[i], InputStr, Offset);
if CurrPos > 0 then
begin
BracketsArray[CurrPos] := i;
Offset := CurrPos + 1;
end;
Until CurrPos = 0;
end; // insert the positions of the opening brackets
for i := Low(CBr) to High(CBr) do
begin
Offset := 1;
Repeat
CurrPos := Pos(CBr[i], InputStr, Offset);
if CurrPos > 0 then
begin
BracketsArray[CurrPos] := i;
Offset := CurrPos + 1;
end;
Until CurrPos = 0;
end; // insert the positions of the closing brackets
Setlength(Stack, 0); // initialize the stack to push/pop the last bracket
for i := 0 to High(BracketsArray) do
case BracketsArray[i] of
Low(OBr) .. High(OBr):
begin
Setlength(Stack, Length(Stack) + 1);
Stack[High(Stack)] := BracketsArray[i];
end; // there is an opening bracket
Low(CBr) .. High(CBr):
begin
if Length(Stack) = 0 then
exit(false); // we can not begin an expression with Closing bracket
if Stack[High(Stack)] <> BracketsArray[i] - 10 then
exit(false) // here we do check if the previous bracket suits the
// closing bracket
else
Setlength(Stack, Length(Stack) - 1); // remove the last opening
// bracket from stack
end;
end;
if Length(Stack) = 0 then
result := true;
end;
也许,我们通过创建字节数组做了额外的工作,但似乎这种方法 i) 更容易理解,ii) 很灵活,因为我们可以更改括号表达式的长度,例如使用和检查 开始
/结束
括号等
已附加
一旦我发现主要问题在于组织文件的 block 读取,我就会在这里给出如何做到这一点的想法:
procedure BlckRead;
var
f: file;
pc, pline: { PChar } PAnsiChar;
Ch: { Char } AnsiChar;
LngthLine, LngthPc: word;
begin
AssignFile(f, 'b:\br.txt'); //open the file
Reset(f, 1);
GetMem(pc, FileSize(f) + 1); //initialize memory blocks
inc(pc, FileSize(f)); //null terminate the string
pc^ := #0;
dec(pc, FileSize(f)); //return the pointer to the beginning of the block
GetMem(pline, FileSize(f)); //not optimal, but here is just an idea.
pline^ := #0;//set termination => length=0
BlockRead(f, pc^, FileSize(f)); // read the whole file
//you can optimize that if you wish,
//add exception catchers etc.
LngthLine := 0; // current pointers' offsets
LngthPc := 0;
repeat
repeat
Ch := pc^;
if (Ch <> #$D) and (Ch <> #$A) and (Ch <> #$0) then
begin // if the symbol is not string-terminating then we append it to pc
pline^ := Ch;
inc(pline);
inc(pc);
inc(LngthPc);
inc(LngthLine);
end
else
begin //otherwise we terminate pc with Chr($0);
pline^ := #0;
inc(LngthPc);
if LngthPc < FileSize(f) then
inc(pc);
end;
until (Ch = Chr($D)) or (Ch = Chr($A)) or (Ch = Chr($0)) or
(LngthPc = FileSize(f));
dec(pline, LngthLine);
if LngthLine > 0 then //or do other outputs
Showmessage(pline + #13#10 + Booltostr(ExpressionIsValid(pline), true));
pline^ := #0; //actually can be skipped but you know your file structure better
LngthLine := 0;
until LngthPc = FileSize(f);
FreeMem(pline); //free the blocks and close the file
dec(pc, FileSize(f) - 1);
FreeMem(pc);
CloseFile(f);
end;
关于performance - 检查括号顺序是否有效,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33849127/