Java | TicTacToe右对角线不起作用

时间:2018-11-02 20:32:45

标签: java arrays 2d diagonal

当前正在用Java玩井字游戏,我有一个checkWin()方法可正确运行4种可能的获胜条件中的3种。我遇到的问题是右对角线。

代码:

public boolean checkWin(String player){
    int row = 0; // Holder to count number of player spots in row
    int d1 = 0; // Holder to count number of player spots in right diag.
    int d2 = 0; // Holder to count number of player spots in left diag.
    int[] column = new int[squares[0].length]; /* Holder to count number
    of player spots in column */

    for(int i = 0; i < size; i++){
        row = 0;
        for(int j = 0; j < size; j++){
            if(null == squares[i][j]){
                continue;
            }
            if(squares[i][j].getText().equals(player)){
                row++; /* If spot at [i][j] equals player, increase row */
                column[j]++; /* If spot at [i][j] equals player, increase 
                col */
                if(i == j){ /* If spot at i is equal to j, increase left 
                    diag */
                    d1++;
                } else if ((size - 1) == i + j){ /* If spot at i + j 
                    equals board size - 1, increase right diag. */
                    d2++;
                }
            }
        }
        if(row == size){ 
            /* 
            if spots in row is equal to size (otherwise, if it fills
            the row, return win
            */
            return true;
        }
    }
    if(size == d1 || size == d2){
        /*
        if spots in either diag is equal to size, return win
        */
        return true;
    } 
    for(int i = 0; i < column.length; i++){
        if(column[i] == size){
            /*
            if column is full of the same player character, return win
            */
            return true;
        }
    }
    /*
    otherwise, return false
    */
    return false;
}

问题部分是:

                else if ((size - 1) == i + j){ /* If spot at i + j 
                    equals board size - 1, increase right diag. */
                    d2++;
                }

以这种方式进行设置的原因是2D阵列的工作原理,因此对于3x3板:

[00] [01] [02]

[10] [11] [12]

[20] [21] [22]

如果i + j = size-1,则求值2 + 0、1 + 1、0 + 2都等于2,如果size = 3,则等于size-1,但是当我运行程序并执行右对角移动,它不会返回真正的胜利值。

任何解决此问题的建议将不胜感激。

1 个答案:

答案 0 :(得分:1)

else if ((size - 1) == i + j)

^仅当其上方的if条件为false时,才进行评估。

if(i == j)

i == 1j == 1时,i == j为真,因此(size - 1) == i + j不被评估。

TLDR:摆脱您的else