Question

我有这个重复的代码，我不确定如何只用一种方法制作它。

public int isWonVertical()
    {
        for (int i = 0; i < columns; i++)
        {
            resetCounter();
            for (int j = 0; j < rows; j++)
            {
                if (raster[j, i] == 1) counterPlayer1++;
                else counterPlayer1 = 0;

                if (raster[j, i] == 2) counterPlayer2++;
                else counterPlayer2 = 0;

                if (counterPlayer1 == tokenStreak) return 1;
                if (counterPlayer2 == tokenStreak) return 2;
            }
        }
        return 0;
    }//isWonVertical


public int isWonHorizontal()
    {
        for (int i = 0; i < rows; i++)
        {
            resetCounter();
            for (int j = 0; j < columns; j++)
            {
                if (raster[i, j] == 1) counterPlayer1++;
                else counterPlayer1 = 0;

                if (raster[i, j] == 2) counterPlayer2++;
                else counterPlayer2 = 0;

                if (counterPlayer1 == tokenStreak) return 1;
                if (counterPlayer2 == tokenStreak) return 2;
            }
        }
        return 0;
    }//isWonHorizontal

返回和 resetCounter（）我可以全部放入1方法。但是我如何确保for循环对于垂直/水平不同。我假设它是给出参数，然后检查我给'垂直'或'水平'作为参数。但我不确定如何让它真正起作用。

谢谢。

Answer 1

public int isWon(DirectionEnum enum)
    {
        int counter1 = enum == DirectionEnum.IsVertical ? columns : rows;
        int counter2 = enum == DirectionEnum.IsHorizontal ? columns: rows; 
        for (int i = 0; i < counter1 ; i++)
        {
            resetCounter();
            for (int j = 0; j < counter2; j++)
            {
                if (raster[i, j] == 1) counterPlayer1++;
                else counterPlayer1 = 0;

                if (raster[i, j] == 2) counterPlayer2++;
                else counterPlayer2 = 0;

                if (counterPlayer1 == tokenStreak) return 1;
                if (counterPlayer2 == tokenStreak) return 2;
            }
        }
        return 0;
    }

Answer 2

他的两个参数如何，一个用于内部数组，一个用于外部数据。然后你的客户端（调用代码）需要决定使用什么作为内部或外部，行或列

public int isWon(outerArray, innerArray)
{
    for (int i = 0; i < outerArray; i++)
    {
        resetCounter();
        for (int j = 0; j < innerArray; j++)
        {
            if (raster[i, j] == 1) counterPlayer1++;
            else counterPlayer1 = 0;

            if (raster[i, j] == 2) counterPlayer2++;
            else counterPlayer2 = 0;

            if (counterPlayer1 == tokenStreak) return 1;
            if (counterPlayer2 == tokenStreak) return 2;
        }
    }
    return 0;
}

重复2种方法制作成1种方法

2 个答案: