TicTacToe赢得NxN板的逻辑

目前的逻辑适用于3×3电路板，因为它是静态的。 如何将其转换为NxN逻辑？

Win逻辑通过添加行和列方块来工作。

/* * To determine a win condition, each square is "tagged" from left * to right, top to bottom, with successive powers of 2. Each cell * thus represents an individual bit in a 9-bit string, and a * player's squares at any given time can be represented as a * unique 9-bit value. A winner can thus be easily determined by * checking whether the player's current 9 bits have covered any * of the eight "three-in-a-row" combinations. * * 273 84 * \ / * 1 | 2 | 4 = 7 * -----+-----+----- * 8 | 16 | 32 = 56 * -----+-----+----- * 64 | 128 | 256 = 448 * ================= * 73 146 292 * */ wins = [7, 56, 448, 73, 146, 292, 273, 84], /* * Returns whether the given score is a winning score. */ win = function (score) { for (var i = 0; i < wins.length; i += 1) { if ((wins[i] & score) === wins[i]) { return true; } } return false; }, 

我的小提琴就在这里

Answer 1

因此，要以编程方式执行此操作，您可以使用类来跟踪每个单元格所处的“设置”，即“row1”或“col1”：

在i / j创建循环中：

 cell.addClass('col' + j); // The cell is in column j cell.addClass('row' + i); // The cell is in row i if (i == j) { cell.addClass('dia0'); // The cell is in the down/right diagonal } if (j == SIZE - i - 1) { cell.addClass('dia1'); // The cell is in the up/right diagonal } 

然后，在win() ，单击最后一个单元格。 对于单元所属的每个类，检查具有包含X （或O ）的类的单元格数是否等于表大小：

 win = function (clicked) { // Get all of the classes this cell belongs to var memberOf = clicked[0].className.split(/\s+/); // Check elements with the same class, and see if they contain "turn", ie X or O for (var i=0; i 

JSFiddle演示

Answer 2

这符合您的三个列表，我已经手动检查了两个。 这当然是正确的，因为代码真的很简单，但你可能想要检查它：

 var getWins = function(size) { var val = 1, cells = [], wins = []; for (var i = 0; i < size; i++) { cells[i] = []; for (var j = 0; j < size; j++) { cells[i][j] = val; val *= 2; } } var rowWins = [], colWins = [], mainDiagWin = 0, antiDiagWin = 0; for (i = 0; i < size; i++) { rowWins[i] = 0; colWins[i] = 0; mainDiagWin += cells[i][i]; antiDiagWin += cells[i][size - i - 1]; for (j = 0; j < size; j++) { rowWins[i] += cells[i][j]; colWins[i] += cells[j][i]; } } return rowWins.concat(colWins, mainDiagWin, antiDiagWin); }; getWins(2); //=> [3, 12, 5, 10, 9, 6] getWins(3); //=> [7, 56, 448, 73, 146, 292, 273, 84] getWins(4) //=> [15, 240, 3840, 61440, 4369, 8738, 17476, 34952, 33825, 4680]