3

这适用于井字游戏，其中网格可以调整为任意数字（3x3 4x4 8x8 等）

当表单加载时，一个方法读取网格大小并使用一维按钮数组中的许多按钮填充表单。该数组称为按钮数组。

仅使用一维 buttonArray 而不使用 LINQ，我如何比较 buttonArray 中按钮的 .Text 值，以查看它们对于 CheckWinner() 函数是“X”还是“O”。

创建按钮网格后，我有一个按钮点击事件处理程序：

private void button_click(object sender, EventArgs e)
    {
        Button b = (Button)sender;
        b.Text = "X";
        b.Enabled = false;
        CheckWinner(buttonArray);
}

然后我调用我的 CheckWinner 函数并传入数组 buttonArray。

同样，我只是在寻找一种不使用 LINQ 来检查/比较值的原始方法。如果我知道网格每一边的长度，我可以说对于水平获胜线，我正在寻找以“X”作为其 .Text 属性的那么多按钮。

因此，如果我有一个 5x5 的网格，并且我在 buttonArray 中有 25 个按钮，我如何从网格的 0 索引开始检查每 5 个的 .Text 值是否为“X”，然后打印一条消息，如果连续 5 个相同，在这种情况下为“X”。

for (int z = 0; z < root; z++) //increments the column to start on
        {
            vCount = 0; //Starts the win counter at zero when moving to the next column
            for (int i = z; i < arrLength; i=i+root) //starts at the first column and increments it by root to test the next button in the column
            {
                string bText = buttonArray[i].Text;
                if (bText == "X")
                    vCount++;
                if (vCount == root)
                {
                    MessageBox.Show("Vertical row winner found !");
                    break;
                }
            }
        }//end of vertical column test

我做了这样的垂直测试？但我认为将它们合二为一会更好。

4

3 回答 3

2

假设您要检查是否 Button.Text == "X" 在 every element of a horizontal row. 以下是一个非常基本的方法......

public void CheckWinner(Buttons[] buttonArray)
{
    int arrLength = buttonArray.Length; 
    int hCount = 0;
    //int vCount = 0;
    int root = (int)Math.Sqrt(Convert.ToDouble(arrLength));  

    for (int i = 0; i < arrLength ; i++)
    { 
        string bText = buttonArray[i].Text;

        hCount = i % root == 0? 0 : hCount;

        if(bText == "X") 
            hCount++;

        if(hCount == root) 
        {
            Console.WriteLine("Horizontal row winner found !");
            break;
        }
    }

}

要检查垂直和/或水平赢家：

public void CheckWinner(Buttons[] buttonArray)
{
    int arrLength = buttonArray.Length; 
    int hCount = 0;
    Int vCount = 0;
    int root = (int)Math.Sqrt(Convert.ToDouble(arrLength));  

    for (int i = 0;  i < root;  i++)
    {
        hCount = 0;
        vCount = 0;
        for(int j = 0;  j < root; j++)
        {
           if(buttonArray[ (i * root) + j ].Text == "X")
              hCount++;

           if(buttonArray[ i + (root * j) ].Text == "X")
              vCount++;
        }

        if( hCount + vCount == 2 * root)
        {
           Console.WriteLine("Horizontal and Virtical winner found !");
           break;
        } 
        else if ( hCount == root )
        { 
           Console.WriteLine("Horizontal winner found !");
           break;
        }
        else if ( vCount == root )
        { 
           Console.WriteLine("Virtical winner found !");
           break;
        }

    }
}
于 2013-03-01T19:01:26.653 回答
2

尽管看起来所有其他答案也都有效，但我会将我的选择权投入到戒指中。我的是一个基本的实现。它只是沿着每一行进行，并检查每一行中是否有相同的值。垂直版本就像重新排列我的两个 for 循环一样简单，对角线也不会太难。

在我看来，这个问题的主要技巧只是展示对数组行和列的理解，以及如何在仅给定一维数组的情况下模拟二维数组。

对于水平：

string CheckWinnerHorizontal(Button[] buttonArray) {
    int N = (int)Math.Sqrt(buttonArray.Length);
    for (int row = 0; row < N; ++row)
    {
        string winner = "";
        for (int col = 0; col < N; ++col)
        {
            string value = buttonArray[row * N + col].Text;
            if (winner == "") { winner = value; }
            else if (winner != value) { winner = "none"; }
        }
        if (winner != "none" && winner != "")
        {
            return winner;
        }
    }
    return "";
于 2013-03-01T19:13:06.860 回答
1

当然，你可以试一试——我会在这里逐行做，by-cols 应该非常相似：

（编辑 - 哈，显然我看不懂......但对其他人来说可能仍然是一个有用的答案......因为我已经证明我没有关注 OP，让我对此进行扩展并填写垂直/对角线...）

（LINQPad 就绪）

void Main()
{
    // Quickish and very dirty way to generate the grid
    var lineLength = 3;
    var rnd = new Random();
    var gridSrc = 
        from r in Enumerable.Range(0, lineLength)
        from c in Enumerable.Range(0, lineLength)
        select new { Row = r, Col = c, Text = rnd.Next(0,2) > 0 ? "X" : "O" };
    var grid = gridSrc.ToArray();

    // ok, now for the query
    var horizontalWinners =
        // need the cell and it's index - this is one way to do that
        from cellTuple in grid.Select((cell, idx) => Tuple.Create(idx, cell))
        let idx = cellTuple.Item1
        let cell = cellTuple.Item2
        // figure out which row its in
        let row = idx / lineLength
        // figure out which column its in
        let col = idx % lineLength
        // for rows, group by row #
        group cell by row into byRow
        // only count if all cells in that row are same
        where byRow.All(rowCell => rowCell.Text == "X") 
             || byRow.All(rowCell => rowCell.Text == "O")
        // tell us what row (and who won)
        select new { byRow.Key, byRow.First().Text };

var verticalWinners =
    from cellTuple in grid.Select((cell, idx) => Tuple.Create(idx, cell))
    let idx = cellTuple.Item1
    let cell = cellTuple.Item2
    let row = idx / lineLength
    let col = idx % lineLength
    group cell by col into byCol
    where byCol.All(colCell => colCell.Text == "X") 
                 || byCol.All(colCell => colCell.Text == "O")
    select new { byCol.Key, byCol.First().Text };

var topLeftBottomRightDiagonalWinners =
    from cellTuple in grid.Select((cell, idx) => Tuple.Create(idx, cell))
    let idx = cellTuple.Item1
    let cell = cellTuple.Item2
    let row = idx / lineLength
    let col = idx % lineLength
    let fwdSlash = (row == col)
    group cell by fwdSlash into byDiag
    where byDiag.Key && byDiag.All(d => d.Text == byDiag.First().Text)
    select new { 
                 Text = byDiag.First().Text, 
                 Pos = string.Join("", byDiag.Select(c => Tuple.Create(c.Col, c.Row).ToString())) 
            };

var topRightBottomLeftDiagonalWinners =
    from cellTuple in grid.Select((cell, idx) => Tuple.Create(idx, cell))
    let idx = cellTuple.Item1
    let cell = cellTuple.Item2
    let row = idx / lineLength
    let col = idx % lineLength
    let backSlash = (row + col) == (lineLength - 1)
    group cell by backSlash into byDiag     
    where byDiag.Key && byDiag.All(d => d.Text == byDiag.First().Text)
    select new { 
               Text = byDiag.First().Text, 
               Pos = string.Join("", byDiag.Select(c => Tuple.Create(c.Col, c.Row).ToString())) 
            };

for(int r=0;r<lineLength;r++)
{
    for(int c=0;c<lineLength;c++)
    {
        Console.Write(grid[r*lineLength+c].Text + " ");
    }
    Console.WriteLine();
}
foreach(var row in horizontalWinners)
{
    Console.WriteLine("{0} wins on row {1}", row.Text, row.Key);
}
foreach(var col in verticalWinners)
{
    Console.WriteLine("{0} wins on col {1}", col.Text, col.Key);
}
foreach (var diag in topLeftBottomRightDiagonalWinners
                .Concat(topRightBottomLeftDiagonalWinners)) 
{
    Console.WriteLine("{0} wins on diagonal {1}", diag.Text, diag.Pos);     
}
}
于 2013-03-01T19:01:17.640 回答