我目前正在使用网页(HTML,CSS,Javascript)编写基本的Tic Tac Toe游戏(多人游戏,没有人工智能)。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();
}
}
}
目前,checkWinner
函数仅检查与winningCombo
数组中数组元素的完全匹配(例如:如果组合为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?希望我的解释有意义,谢谢。
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;
}
}
答案 0 :(得分:0)
在与winsCombo比较之前对播放的数组进行排序。
playedComboO.sort(function (a, b) {
return a - b;
});
我认为playedComboO.sort()
(和playedComboX.sort()
)也可以使用,但默认的排序功能是字符串而不是数字。
答案 1 :(得分:0)
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;
}
}
}
}