Question

我回答了一个问题。经过大量的代码重写，我的tic tac toe看起来很好，除了一个明显的问题。当我尝试玩时，胜利条件似乎随机激活 - 我将x-o-x排成一行并说“玩家X赢了！”但是当我有o - o - o时它不会检测到赢得。尽管画得好。

赢取条件代码：

 // win conditions. if true, set win==true; else set win==false
            if (square[0].getText().equals(square[1].getText())
                    && square[1].getText().equals(square[2].getText())
                    != square[0].getText().isEmpty()) {
                    win = true;}

            if (square[3].getText().equals(square[4].getText())
                    && square[4].getText().equals(square[5].getText())
                    != square[3].getText().isEmpty()) {
                    win = true;}

            if (square[6].getText().equals(square[7].getText())
                    && square[7].getText().equals(square[8].getText())
                    != square[6].getText().isEmpty()) {
                    win = true;}

            if (square[0].getText().equals(square[3].getText())
                    && square[3].getText().equals(square[6].getText())
                    != square[0].getText().isEmpty()) {
                    win = true;}

            if (square[1].getText().equals(square[4].getText())
                    && square[4].getText().equals(square[7].getText())
                    != square[1].getText().isEmpty()) {
                    win = true;}

            if (square[2].getText().equals(square[5].getText())
                    && square[5].getText().equals(square[8].getText())
                    != square[2].getText().isEmpty()) {
                    win = true;}

            if (square[0].getText().equals(square[4].getText())
                    && square[4].getText().equals(square[8].getText())
                    != square[0].getText().isEmpty()) {
                    win = true;}

            if (square[6].getText().equals(square[4].getText())
                    && square[4].getText().equals(square[2].getText())
                    != square[6].getText().isEmpty()) {
                    win = true;}

            else{win = false;
            }

动作代码：

  public void actionPerformed (ActionEvent e) {
            //one more move has gone by, calculate player turn + player letter
            move++;
            if (move % 2 == 0) {
                    player = 1; letter = "X";
            }else{
                    player = 2; letter = "O";
            }

            playergo.setText("It is player " + player + "'s go!");

            //set square letter to player's letter, disable square so no further moves can be made there
            for (int i=0; i<=8; i++){
                    if (e.getSource() == square[i]){
                            square[i].setText(letter);
                            square[i].setEnabled(false);
                    }
            }

完整的粘贴代码here。提前谢谢大家!!!

Answer 1

由于您未在胜利条件下使用if / else，因此您将始终检查上一个if，如果该结算为false，则会设置win = false

Answer 2

你可以这样做（在伪代码中）

int winPositions[][] = { {0,1,2}, {3,4,5}, {6,7,8}, //horizontal
                         {0,3,6}, {1,4,7}, {2,5,8}, //vertical
                         {0,4,8}, {2,4,6} }; //diagnol 

for i = 0 to winPosition.size
  int positions[] = winPosition[i]; //know its size is 3
  bool hasWon = board[ positions[i] ] == board[ positions[i+1] ] && board[positions[i+1]] == board[positions[i+2]];
  if(hasWon) return true;

Tic Tac Toe赢得未正确激活的条件

2 个答案: