Question

我有一个关联数组，我按升序排序，现在我想对这些值做出一些决定。我不明白我该怎么办。

让我解释一下我想要实现的目标：

假设：我有以下原始数组：

Array
(
    [0] => Array
        (
            [cell] => b2
            [sign] => &#10008;
        )    
    [1] => Array
        (
            [cell] => a1
            [sign] => &#9711;
        )
)

然后我把它分类为：

Array
(
    [0] => Array
        (
            [cell] => a1
            [sign] => &#9711;
        )
    [1] => Array
        (
            [cell] => b2
            [sign] => &#10008;
        )
)

现在我想做出决定，例如细胞a1 = b1 = c1或a1 = b2 = c3等的值。现在我该怎么做？

Answer 1

猜测你的方格只有 3x3 ，我想说最简单的方法是将获胜的匹配存储在一个数组中，然后使用array_intersect()比较两个数组来比较它们并返回数组中的重复值。要获得重复数量，只需打包count()函数并比较玩家的一个或两个值是否与$winArr匹配。显然，通过停止输入数组的重复值仍然可以做到，在每次输入后循环foreach()等等......但至少你的逻辑是正确的。

$winner; //Winner --> Used when game finished

$winArr = [

    ['c1','c2','c3'], 
    ['b1','b2','b3'], // --> Vertical Win
    ['a1','a2','a3'],

    ['a1','b2','c3'],
    ['c1','b2','c3'], // --> Diagonal Win

    ['a1','b1','c1'],
    ['a2','b2','c2'], // --> Horizontal Win
    ['a3','b3','c3'],

];

$gameArr = [ // --> `Input Values Array = Dimension for P1 & P2 for ease

'playerOne' => ['a1', 'b1', 'c2', 'c1'],
'playerTwo' => ['a3', 'b3', 'a2'],    

];



foreach ($winArr as $k=>$v) {

    if (count(array_intersect($v, $gameArr['playerOne'])) === 3) {

        $winner = 'Player One!';       
        break;

    } elseif(count(array_intersect($v, $gameArr['playerTwo'])) === 3) {

        $winner = 'Player Two!';
        break;

    }

}

if (isset($winner)) {

    echo $winner;

}

输出

Player One!

请参阅： http://www.php.net/manual/en/function.array-intersect.php

Answer 2

我建议用零基索引来指代细胞的位置。它可以很容易地使用数学技巧，如模数或整数除法来选择行或列。否则，您必须使用大量偏移来避免一个一个错误。

IE，董事会可以贴上标签：

0 1 2
3 4 5
6 7 8

游戏状态可以用字符数组表示，字符' - '表示空，x和o表示x和o在那里播放。

这是一个人类游戏视图，游戏存储在一个字符串数组中：

  x
o x o
o x

var state = "-x-oxoox-";

要检查游戏是否已获胜，请发挥功能。

如果没有人赢，该函数返回' - '，如果赢了则返回'x'，如果赢了则返回'o'。

我将启动一个硬编码每个路径进行检查的实现。然后我将展示如何生成要检查的路径。

function WhoIsWinningHardCode(state) {

  var rows     = IsWin(state, [[0,1,2],[3,4,5],[6,7,8]]);
  var cols     = IsWin(state, [[0,3,6],[1,4,7],[2,5,8]]);
  var diagUp   = IsWin(state, [[6,4,2]]);
  var diagDown = IsWin(state, [[0,4,8]]);

  if (rows     !== '-') return rows;
  if (cols     !== '-') return cols;
  if (diagUp   !== '-') return diagUp;
  return diagDown;

}

function WhoIsWinningGeneratePaths(state) {

  var rows     = IsWin(state, GenPaths(3, 0, 1, 3));
  var cols     = IsWin(state, GenPaths(3, 0, 3, 1));
  var diagUp   = IsWin(state, GenPaths(1, 2, 2, 0));
  var diagDown = IsWin(state, GenPaths(1, 0, 4, 0));

  if (rows     !== '-') return rows;
  if (cols     !== '-') return cols;
  if (diagUp   !== '-') return diagUp;
  return diagDown;

}

function GenPaths(count, start, incrementA, incrementB) {
   var paths = [];
   for (var i = 0; i < count; i++) {
      var path = [];
      for (var j = 0; j < 3; j++) {
         path.push(start + i*incrementB + j*incrementA)
      }
      paths.push(path);
   }
   return paths;
}

function IsWin(state, paths) {
  for (var i = 0; i < paths.length; i++) {
     var path = paths[i];
     var a = state[path[0]];
     var b = state[path[1]];
     var c = state[path[2]];
     if (a !== '-')
       if (a == b)
         if (a == c)
           return a;
  }

  return '-';
}

Answer 3

这是Javascript中更通用的实现。它仅在2x2和3x3游戏状态下进行测试，可能需要针对其他游戏尺寸进行调试：

＆＃13;

// xxx
// oo-
// o--  
alert("game a: " + WhoIsWinningGeneratePaths('xxxoo-o--'));


// x-x
// oo-
// o--  
alert("game b: " + WhoIsWinningGeneratePaths('x-xoo-o--'));

// x-x
// ooo
// ---  
alert("game c: " + WhoIsWinningGeneratePaths('x-xooo---'));

// x-
// oo  
alert("game d: " + WhoIsWinningGeneratePaths('x-oo'));

// x-
// ox  
alert("game e: " + WhoIsWinningGeneratePaths('x-ox'));


// x-
// o-  
alert("game f: " + WhoIsWinningGeneratePaths('x-o-'));


function WhoIsWinningGeneratePaths(state) {

  var n = Math.sqrt(state.length);

  var rows = IsWin(state, GenPaths(n, 0,     1,   n, n));
  var cols = IsWin(state, GenPaths(n, 0,     n,   1, n));
  var diUp = IsWin(state, GenPaths(1, n-1, n-1,   0, n));
  var diDn = IsWin(state, GenPaths(1, 0,   n+1,   0, n));

  if (rows !== '-') return rows;
  if (cols !== '-') return cols;
  if (diUp !== '-') return diUp;
  return diDn;

}

function GenPaths(count, start, incrementA, incrementB, lengthToWin) {
  var paths = [];
  for (var i = 0; i < count; i++) {
    var path = [];
    for (var j = 0; j < lengthToWin; j++) {
      path.push(start + i * incrementB + j * incrementA)
    }
    paths.push(path);
  }
  return paths;
}


function IsWin(state, paths) {
  for (var i = 0; i < paths.length; i++) {
    var currentPathResult = IsPathWin(state, paths[i]);
    if (currentPathResult != '-')
      return currentPathResult;
  }
  return '-';
}

function IsPathWin(state, path) {
  var first = state[path[0]];
  for (var j = 1; j < path.length; j++)
  {
    var compareToFirst = state[path[j]];
    if (compareToFirst !== first)
      return '-';
  }
  return first;
}

＆＃13;

根据数组中的值做出决策

3 个答案: