5

（在 2 人游戏中）查看井字游戏获胜的最佳方式是什么？现在我正在使用类似于以下内容的东西：

if (btnOne.Text == "X" && btnTwo.Text == "X" && btnThree.Text == "X")
{
    MessageBox.Show("X has won!", "X won!");
    return;
}
else
// I'm not going to write the rest but it's really just a bunch
// if statements.

那么如何摆脱多重if呢？

4

4 回答 4

11

一些东西：

rowSum == 3 || columnSum == 3 || diagnolSum == 3

.. ?

于 2009-04-11T17:25:55.600 回答
3

如果将按钮存储在多维数组中，则可以编写一些扩展方法来获取行、列和对角线。

public static class MultiDimensionalArrayExtensions
{
  public static IEnumerable<T> Row<T>(this T[,] array, int row)
  {
    var columnLower = array.GetLowerBound(1);
    var columnUpper = array.GetUpperBound(1);

    for (int i = columnLower; i <= columnUpper; i++)
    {
      yield return array[row, i];
    }
  }

  public static IEnumerable<T> Column<T>(this T[,] array, int column)
  {
    var rowLower = array.GetLowerBound(0);
    var rowUpper = array.GetUpperBound(0);

    for (int i = rowLower; i <= rowUpper; i++)
    {
      yield return array[i, column];
    }
  }

  public static IEnumerable<T> Diagonal<T>(this T[,] array,
                                           DiagonalDirection direction)
  {
    var rowLower = array.GetLowerBound(0);
    var rowUpper = array.GetUpperBound(0);
    var columnLower = array.GetLowerBound(1);
    var columnUpper = array.GetUpperBound(1);

    for (int row = rowLower, column = columnLower;
         row <= rowUpper && column <= columnUpper;
         row++, column++)
   {
      int realColumn = column;
      if (direction == DiagonalDirection.DownLeft)
        realColumn = columnUpper - columnLower - column;

      yield return array[row, realColumn];
    }
  }

  public enum DiagonalDirection
  {
    DownRight,
    DownLeft
  }
}

如果您使用TableLayoutPanel3 行和 3 列的 a，您可以轻松地以可编程方式创建按钮并将其存储到Button[3, 3]数组中。

Button[,] gameButtons = new Button[3, 3];

for (int row = 0; column <= 3; row++)
  for (int column = 0; column <= 3; column++)
  {
    Button button = new Button();
    // button...
    gameLayoutPanel.Items.Add(button);
    gameButtons[row, column] = button;
  }

并检查获胜者：

string player = "X";
Func<Button, bool> playerWin = b => b.Value == player;
gameButtons.Row(0).All(playerWin) ||
// ...
gameButtons.Column(0).All(playerWin) ||
// ...
gameButtons.Diagonal(DiagonalDirection.DownRight).All(playerWin) ||
// ...
于 2009-04-11T18:24:06.087 回答
2

另一种简单的方法是将可获胜的位置保存为数组中的数据，并使用循环来检查所有可能的获胜条件，而不是使用多个 ifs 语句

// winnable positions
var winnables = new[] {
    "012",
    "345",
    "678",
    "036",
    "147",
    "258",
    "048",
    "246"
};

// extracted from btnOne Two Three....
var gameState = new[] { "X", "O", "X", "whatever" };


string winner = null;

// check each winnable positions
foreach (var position in winnables) {

    var pos1 = int.Parse(position[0].ToString());
    var pos2 = int.Parse(position[1].ToString());
    var pos3 = int.Parse(position[2].ToString());

    if (gameState[pos1] == gameState[pos2] &&
        gameState[pos2] == gameState[pos3])
        winner = gameState[pos1];

}

// do we have a winner?
if (!string.IsNullOrEmpty(winner))
    /* we've got a winner */

基本上，不要使用 btnOne btnTwo btnThree，使用适当的 Button 数组或以更易于访问的格式保存游戏状态的数组，这样会更容易计算。

于 2009-04-11T17:42:45.590 回答
0

我倾向于做这样的事情：

bool x_wins =
    Enumerable
        .Range(0, 3)
        .SelectMany(i => new Func<int, string>[] { x => array[i, x], x => array[x, i] })
        .Concat(new Func<int, string>[] { x => array[x, x], x => array[2 - x, x], })
        .Where(f => String.Concat(Enumerable.Range(0, 3).Select(x => f(x))) == "XXX")
        .Any();
于 2021-04-22T00:21:15.190 回答