2

我目前正在使用网络（HTML、CSS、Javascript）编写一个基本的井字游戏（多人游戏，无 AI）。游戏逻辑显然都在 Javascript 中。我有个问题。以下是我的代码。

window.onload = function () {
    console.log("Page has loaded");
}
var ctx;
var turn = 0;
var winningCombo = [
    [1, 4, 7],
    [2, 5, 8],
    [3, 6, 9],
    [1, 2, 3],
    [4, 5, 6],
    [7, 8, 9],
    [1, 5, 9],
    [3, 5, 7]
];
var playedComboX = [];
var playedComboO = [];
var filledSquares = [0];
var filled = false;
console.log(winningCombo.length);
var checkWinnerX = function () {
    for(var i = 0; i < winningCombo.length; i++) {
        console.log('Its in the X check winner loop');
        if((winningCombo[i][0] == playedComboX[0]) && (winningCombo[i][1] == playedComboX[1]) && (winningCombo[i][2] == playedComboX[2])) {
            alert("Congrats, you have won!");
            return true;
        }
        return false;
    }
}
var checkWinnerO = function () {
    for(var i = 0; i < winningCombo.length; i++) {
        console.log('Its in the 0 check winner loop');
        if(winningCombo[i][0] == playedComboO[0] && winningCombo[i][1] == playedComboO[1] && winningCombo[i][2] == playedComboO[2]) {
            console.log('It has passed the if statement for X');
            alert("Congrats, you have won!");
            return true;
        }
        return false;
    }
}
var checkWinner = function () {}
var draw = function (squareNumber) {
    console.log('draw has been called');
    var squareID = "square" + squareNumber;
    var squareClicked = document.getElementById(squareID);
    ctx = squareClicked.getContext('2d');
    for(var i = 0; i < filledSquares.length; i++) {
        if(filledSquares[i] == squareNumber) {
            filled = true;
        } else {
            filled = false;
        }
    }
    if(filled == true) {
        alert("Invalid Move! Square is already occupied");
    } else {
        filledSquares.push(squareNumber);
        ctx.beginPath();
        if(turn % 2 == 0) {
            //Drawing a 'X'
            ctx.moveTo(20, 20);
            ctx.lineTo(135, 135);
            ctx.moveTo(135, 20);
            ctx.lineTo(20, 135);
            ctx.lineWidth = 6;
            ctx.stroke();
            turn++;
            playedComboX.push(squareNumber);
            checkWinnerX();
        } else {
            //Drawing a Circle
            ctx.arc(75, 75, 65, 2 * Math.PI, false);
            ctx.lineWidth = 6;
            ctx.stroke();
            turn++;
            playedComboO.push(squareNumber);
            checkWinnerO();
        }
    }
}

现在，这些checkWinner函数只检查与数组中的数组元素是否完全匹配winningCombo（例如：如果组合是3,1,2而不是，1,2,3则它不会注册）。无论如何我可以检查每个元素中3个数字的所有可能组合吗？希望我的解释有道理，谢谢。

PS：如果任何代码写得不如我好，请原谅我，这是我第一次尝试编写游戏。但要随心所欲！

更新 2：

var checkWinnerX = function () {
    for(var i = 0; i < winningCombo.length; i++) {
        console.log('Its in the X check winner loop');
        if((winningCombo[i][0] == sortedArrayX[0]) && (winningCombo[i][1] == sortedArrayX[1]) && (winningCombo[i][2] == sortedArrayX[2])) {
            alert("Congrats, you have won!");
            return true;
        }
        return false;
    }
}
4

2 回答 2

1

我认为更好的方法是：

for(var i = 0; i < winningCombo.length; i++) {
    if ( playedComboO.indexOf(winningCombo[i][0]) >= 0 ) {
        if ( playedComboO.indexOf(winningCombo[i][1]) >= 0 ) {
            if ( playedComboO.indexOf(winningCombo[i][2]) >= 0 ) {
                 //blah blah blah you win whatever
            alert("Congrats, you have won!");
        return true;
            }
       }
    }
}
于 2013-08-19T18:25:35.497 回答
0

在与 winsCombo 比较之前对已播放的数组进行排序。

playedComboO.sort(function (a, b) {
    return a - b;
});

我认为playedComboO.sort()(and playedComboX.sort()) 也可以工作，但默认排序功能是按字符串而不是按数字。

于 2013-03-28T00:47:13.820 回答