Question

我有以下测试多维数组，它反映了标准的Connect Four游戏板：

tree['left']['left'] == None

我正在编写一个js验证，它会检查电路板阵列中的每个元素，以便在横向，向下或对角线上进行4种类型的匹配。查看上面的测试板，我的验证器应该返回var board = [['-','-','-','-','-','-','-'], ['-','-','-','-','-','-','-'], ['-','-','-','R','R','R','R'], ['-','-','-','Y','Y','R','Y'], ['-','-','-','Y','R','Y','Y'], ['-','-','Y','Y','R','R','R']];作为正确答案，因为第三行中有4个R元素的连接匹配。这是我的代码：

所以我基本上使用function fourTogether(a,b,c,d) { return (a != '-') && (a == b) && (a == c) && (a == d); } function connectFour(board) { // check RIGHT for (row=col=0; row<3 && col<7; row++, col++) { if (fourTogether(board[row][col], board[row+1][col], board[row+2][col], board[row+3][col])) { return board[row][col]; } } // check DOWN for (row=col= 0; row<6 && col<4; row++, col++) { if (fourTogether(board[row][col], board[row][col+1], board[row][col+2], board[row][col+3])) { return board[row][col]; } } // check DIAGONAL to RIGHT for (row=col=0; row<3 && col<4; row++, col++) { if (fourTogether(board[row][col], board[row+1][col+1], board[row+2][col+2], board[row+3][col+3])) { return board[row][col]; } } // check DIAGONAL to LEFT for (row=3, col=0; row<6 && col<4; row++, col++) { if (fourTogether(board[row][col], board[row-1][col+1], board[row-2][col+2], board[row-3][col+3])) { return board[row][col]; } } return board.indexOf('-') > -1 ? 'in progress' : 'draw'; }函数来建立4-of-kind匹配并省略fourTogether元素（代表未填充的boardlots）。然后在'-'函数中，我使用四个不同的connectFour循环来迭代整个数组，具体取决于检查的方向，然后我添加FOR语句来检查IF对递增的板元素起作用。最后，最后还有一个三元语句应该返回＆＃34; draw＆＃34;或者＆＃34;正在进行的游戏＆＃34;如果电路板中没有4种匹配，则取决于是否存在空fourTogether电路板。不幸的是，我正在返回'-'，而且我对于代码失败的具体位置感到有些不知所措。任何帮助或指示都将非常感激。

Answer 1

for循环，如：

for (row=col=0; row<3 && col<7; row++, col++) {

将迭代(0,0),(1,1),(2,2)。你想要的是(0,0),(0,1),(0,2),...,(1,0),(1,1),(1,2),...。要实现这一点，您必须使用嵌套循环：

for (row = 0; row<3; row++) {
    for (col = 0; col<7; col++) {
        // do that check
    }
}

Answer 2

如果必须，请使用两个嵌套循环并检查四个方向中的每一个。

＆＃13;

var board = [['-','-','-','-','-','-','-'],
             ['-','-','-','-','-','-','-'],
             ['-','-','-','R','R','R','R'],
             ['-','-','-','Y','Y','R','Y'],
             ['-','-','-','Y','R','Y','Y'],
             ['-','-','Y','Y','R','R','R']];

function fourTogether(a,b,c,d) {
   return (a != '-') && (a == b) && (a == c) && (a == d);
}

function connectFour(board) {
  var bl = board.length, bw = board[0].length;

  // loop through the whole board once not a bunch of times
  for (var row = 0; row < bl; row++) {
    for (var col = 0; col < bw; col++) {
      var sq = board[row][col];
      
      // check right if we have to
      if (col < bw - 3 &&
          fourTogether(sq, board[row][col+1], board[row][col+2], board[row][col+3])) {
        return sq;
      }
      // check down if we have to
      if (row < bl - 3 &&
          fourTogether(sq, board[row+1][col], board[row+2][col], board[row+3][col])) {
        return sq;
      }
      // down right
      if (row < bl - 3 && col < bw - 3 &&
          fourTogether(sq, board[row+1][col+1], board[row+2][col+2], board[row+3][col+3])) {
        return sq;
      }
      // down left
      if (row < bl - 3 && col > 2 && 
          fourTogether(sq, board[row+1][col-1], board[row+2][col-2], board[row+3][col-3])) {
        return sq;
      }
    }      
  } 
  //board.indexOf('-') > -1 ? return 'in progress' : return 'draw'; //?????
  return "no winner";
}
alert(connectFour(board));

＆＃13;

Answer 3

我会对你解决这个问题。

首先，设置功能以轻松获取测试线

function row(board, i) {
    return board[i].join('');
}
function col(board, j) {
    return board.map(e => e[j]).join('');
}
function diagDown(board, i) {
    return board.map((e, j) => e[i - board.length + j] || '').join('');
}
function diagUp(board, i) {
    return board.slice(0).reverse().map((e, j) => e[i - board.length + j] || '').join('');
}

（如果你想了解对角线的内容，请尝试在控制台中使用一些值进行操作，并了解它是如何进行映射的）

现在迭代有效行

function whoWon(board) {
    var i, s, r = 'RRRR', y = 'YYYY';
    // rows
    for (i = 0; i < board.length; ++i) {
        s = row(board, i);
        if (s.indexOf(r)) return 'R';
        if (s.indexOf(y)) return 'Y';
    }
    // cols
    for (i = 0; i < board[0].length; ++i) {
        s = col(board, i);
        if (s.indexOf(r)) return 'R';
        if (s.indexOf(y)) return 'Y';
    }
    // diagonals
    for (i = 4; i <= board.length + board[0].length - 4; ++i) {
        s = diagDown(board, i);
        if (s.indexOf(r)) return 'R';
        if (s.indexOf(y)) return 'Y';
        s = diagUp(board, i);
        if (s.indexOf(r)) return 'R';
        if (s.indexOf(y)) return 'Y';
    }
    return '-';
}

现在有

whoWon(board); // "R"

另请注意

function isDraw(board) {
    return board[0].every(e => e !== '-');
}

Javascript - 连接四个游戏验证

3 个答案: