Javascript - Tic Tac Toe - 如何遍历获胜条件？

Question

我正在编写一个可以正常运行的井字游戏程序，我已经差不多完成了。除了我的胜利条件太长而且丑陋。看起来像这样。

function checkWin(){
  if(board[0].textContent === "X" &&
    board[1].textContent === "X" &&
    board[2].textContent === "X"
  ) { alert("Win")}
  else if (
    board[3].textContent === "X" &&
    board[4].textContent === "X" &&
    board[5].textContent === "X"
  ) { alert("Win")}
   else if (
    board[6].textContent === "X" &&
    board[7].textContent === "X" &&
    board[8].textContent === "X"
  ) { alert("Win")}

}

我只写了几个获胜条件，因为如果我写了整个东西，它会更长。我想知道如何编写一个较短的版本。我正在考虑做一个数组并循环遍历它，但我不知道如何做。它将是这样的。

var winConditions = [[0,1,2], [3,4,5], [6,7,8], [0,3,6], [1,4,7], [2, 5,8], [0,4,8], [6,4,2]]

但是我如何使用 checkWin 遍历 winConditions 并使其等于 X 和 O？

Answer 1

使用 ES6

function checkWin(player){
  // player = 'X' or 'O'

  const horizontal = [0,3,6].map(i=>{return[i,i+1,i+2]});
  const vertical = [0,1,2].map(i=>{return[i,i+3,i+6]});
  const diagonal = [[0,4,8],[2,4,6]];

  var allwins = [].concat(horizontal).concat(vertical).concat(diagonal);
  
  let res = allwins.some(indices => { 
  return board[indices[0]] == player && board[indices[1]] == player && board[indices[2]] == player})
  return res;
}
// O O O
// X X O
// X X O
var board=["X","X","O","X","X","O","O","O","O"];
console.log(checkWin("X"),"Expect: false"); // false
console.log(checkWin("O"),"Expect: true"); // true

// negative test case
var board = [];
console.log(checkWin("X"),"Expect: false"); // false

// wierd case of only 1 player
// X X X
var board = ["X","X","X"]
console.log(checkWin("X"),"Expect: true"); //true