arrays - 二维数组的唯一编号分配

Question

我试图将 R*C/2-1 唯一数字写入数组，但随机函数多次给出相同的值。应在随机位置将每个数字填充 2 次。

喜欢:

 ( 1 2 3 4 )
 ( 5 6 7 8 )
 ( 1 2 3 4 )
 ( 5 6 7 8 )

到目前为止我的代码:

...

R=4
C=4

...

 var
    somearray : array [0 .. (C- 1), 0 .. (R- 1)] of integer;

...

    for Row := 0 to (R - 1) do
     for Col := 0 to (C - 1) do
      begin
        somearray [Col, Row] := RandomRange(0, 9);
       end;

...

编辑1:

第二个数组

...

var
   maxnum: array [0 .. (C * R) div 2 - 1] of integer;


max := (C * R) div 2;
for i := 0 to max - 1 do
  maxnum[i] := i;

...

maxnum 的数字为 0..7/1-8

Answer 1

我认为你有两种方法可以做到这一点:

随机位置

迭代这些数字，然后将每个数字放置在随机位置。如果您遇到冲突位置，那么您必须有一些逻辑来找到该值的下一个位置。

使用您的代码我们会得到:

// Initialize the array
for Row := 0 to (R - 1) do
  for Col := 0 to (C - 1) do
    somearray[Col, Row] := 0;
// Populate the random numbers
for randomCount := 1 to 2 do
  for randomNumber := 1 to ((R * C) div 2) do
  begin
    randomPosition := Random(R * C);
    while somearray[randomPosition div C, randomPosition mod C] <> 0 do
    begin
      Inc(randomPosition);
      if randomPosition >= (R * C) then
        randomPosition := 0;
    end;
    somearray[randomPosition div C, randomPosition mod C] := randomNumber;
  end;

将其放在更通用的形式中:

type
  T2DArray = array of array of Integer;

procedure RandomPopulate1(aTheArray: T2DArray);
var
  col: Integer;
  colCount: Integer;
  randomCount: Integer;
  randomNumber: Integer;
  randomPosition: Integer;
  row: Integer;
  rowCount: Integer;
begin
  // Initialize the array
  colCount := Length(aTheArray);
  if colCount = 0 then
    Exit;
  rowCount := Length(aTheArray[0]);
  for col := 0 to colCount - 1 do
    for row := 0 to rowCount - 1 do
      aTheArray[col, row] := 0;
  // Populate the random numbers
  for randomCount := 1 to 2 do
    for randomNumber := 1 to ((colCount * rowCount) div 2) do
    begin
      randomPosition := Random(colCount * rowCount);
      while aTheArray[randomPosition div colCount, randomPosition mod colCount] <> 0 do
      begin
        Inc(randomPosition);
        if randomPosition >= (colCount * rowCount) then
          randomPosition := 0;
      end;
      aTheArray[randomPosition div colCount, randomPosition mod colCount] := randomNumber;
    end;
end;

要使用这个:

var
  somearray: T2DArray;
begin
  Randomize;
  SetLength(somearray, 4, 4);
  RandomPopulate1(somearray);
end;

使用随机位置改变

您可以设置数组中的值，然后随机化位置。因此执行多次随机交换。这是更简单的选择。

// Initialize the array
for Row := 0 to (R - 1) do
  for Col := 0 to (C - 1) do
    somearray[Col, Row] := ((Col * C + Row) div 2) + 1;
// Now randomize the positions
for randomLoop := 0 to (R * C) - 1 do
begin
  randomPosition := Random(R * C);
  randomNumber := somearray[randomLoop div C, randomLoop mod C];
  somearray[randomLoop div C, randomLoop mod C] := somearray[randomPosition  div C, randomPosition  mod C];
  somearray[randomPosition  div C, randomPosition  mod C] := randomNumber;
end;

和以前一样，我们可以有一个更通用的版本:

procedure RandomPopulate2(aTheArray: T2DArray);
var
  col: Integer;
  colCount: Integer;
  randomLoop: Integer;
  randomNumber: Integer;
  randomPosition: Integer;
  row: Integer;
  rowCount: Integer;
begin
  // Initialize the array
  colCount := Length(aTheArray);
  if colCount = 0 then
    Exit;
  rowCount := Length(aTheArray[0]);
  for col := 0 to colCount - 1 do
    for row := 0 to rowCount - 1 do
      aTheArray[col, row] := ((col * colCount + row) div 2) + 1;
  // Now randomize the positions
  for randomLoop := 0 to ((colCount * rowCount) div 2) do
  begin
    randomPosition := Random(colCount * rowCount);
    randomNumber := aTheArray[randomLoop div colCount, randomLoop mod colCount];
    aTheArray[randomLoop div colCount, randomLoop mod colCount] :=
        aTheArray[randomPosition div colCount, randomPosition mod colCount];
    aTheArray[randomPosition div colCount, randomPosition mod colCount] := randomNumber;
  end;
end;