如何编写函数判断井字棋二维数组的获胜方？

需求说明 #

需要编写一个通用函数，接收井字棋二维数组（用"X"、"O"表示玩家，"-"表示空位），判断是否出现获胜局面，并在控制台输出胜负结果——比如传入diagonalWin时，控制台应打印"O Won and X Lost"（注：你给出的diagonalWin示例是O的对角线获胜）。

示例测试数组：

const rowWin = [
  ["O", "O", "O"],
  ["-", "-", "-"],
  ["-", "-", "-"]
];
const colWin = [
  ["-", "X", "-"],
  ["-", "X", "-"],
  ["-", "X", "-"],
];
const diagonalWin = [
  ["-", "-", "O"],
  ["-", "O", "-"],
  ["O", "-", "-"],
];
const diagonalWinInverse = [
  ["X", "-", "-"],
  ["-", "X", "-"],
  ["-", "-", "X"],
];

硬编码方案的问题 #

你当前的代码只能针对特定数组做判断，不具备通用性，且逻辑判断有误：

• 仅针对固定的行、列、对角线位置写死判断，无法覆盖所有可能的获胜情况（比如第二行全X的情况就检测不到）
• 用&&仅检查元素是否存在，没有验证三个元素是否完全相同且不是空位"-"

你的硬编码代码：

if(rowWin[0][0] && rowWin[0][1] && rowWin[0][2]){
    console.log("O wins, X lost")
}else console.log("X wins")

if(colWin[0][1] && colWin[1][1] && colWin[2][1]){
    console.log("X wins, O lost")
}else console.log("O wins")

if(diagonalWin[0][2] && diagonalWin[1][1] && diagonalWin[2][0]){
    console.log("O wins, X lost")
}else console.log("X wins")

if(diagonalWinInverse[0][0] && diagonalWinInverse[1][1] && diagonalWinInverse[2][2]){
    console.log("X wins, X lost")
}else console.log("O wins")

通用解决方案 #

实现思路 #

1. 遍历所有可能的获胜线路：3行、3列、2条对角线
2. 对每条线路，检查三个位置的元素是否完全相同，且不是空位"-"
3. 找到获胜玩家后，直接输出胜负结果；若所有线路都无获胜情况，输出“无胜负”

完整代码 #

function checkTicTacToe(board) {
  // 检查行
  for (let row = 0; row < 3; row++) {
    const cell1 = board[row][0];
    const cell2 = board[row][1];
    const cell3 = board[row][2];
    if (cell1 !== "-" && cell1 === cell2 && cell2 === cell3) {
      const loser = cell1 === "X" ? "O" : "X";
      console.log(`${cell1} Won and ${loser} Lost`);
      return;
    }
  }

  // 检查列
  for (let col = 0; col < 3; col++) {
    const cell1 = board[0][col];
    const cell2 = board[1][col];
    const cell3 = board[2][col];
    if (cell1 !== "-" && cell1 === cell2 && cell2 === cell3) {
      const loser = cell1 === "X" ? "O" : "X";
      console.log(`${cell1} Won and ${loser} Lost`);
      return;
    }
  }

  // 检查主对角线（左上到右下）
  const diag1Cell1 = board[0][0];
  const diag1Cell2 = board[1][1];
  const diag1Cell3 = board[2][2];
  if (diag1Cell1 !== "-" && diag1Cell1 === diag1Cell2 && diag1Cell2 === diag1Cell3) {
    const loser = diag1Cell1 === "X" ? "O" : "X";
    console.log(`${diag1Cell1} Won and ${loser} Lost`);
    return;
  }

  // 检查副对角线（右上到左下）
  const diag2Cell1 = board[0][2];
  const diag2Cell2 = board[1][1];
  const diag2Cell3 = board[2][0];
  if (diag2Cell1 !== "-" && diag2Cell1 === diag2Cell2 && diag2Cell2 === diag2Cell3) {
    const loser = diag2Cell1 === "X" ? "O" : "X";
    console.log(`${diag2Cell1} Won and ${loser} Lost`);
    return;
  }

  // 无获胜情况
  console.log("无胜负");
}

// 测试示例
checkTicTacToe(rowWin); // 输出：O Won and X Lost
checkTicTacToe(colWin); // 输出：X Won and O Lost
checkTicTacToe(diagonalWin); // 输出：O Won and X Lost
checkTicTacToe(diagonalWinInverse); // 输出：X Won and O Lost

代码解释 #

• 遍历行时，逐行检查三个单元格是否一致且非空位
• 遍历列时，逐列检查三个单元格是否一致且非空位
• 对角线单独检查两条特殊线路
• 找到获胜者后立即返回，避免多余判断
• 通过三元运算符快速确定败方，简化逻辑

内容的提问来源于stack exchange，提问作者Ruben