简体   繁体   English

用于检查宾果的 C# 宾果算法

[英]C# Bingo Algorithm for Checking BIngo

I have a 5x5 2D array of type Boolean and I want to check for vertical, horizontal, and diagonal bingo.我有一个布尔类型的 5x5 2D 数组,我想检查垂直、水平和对角宾果游戏。

If values at those specific cells are set to "true", I get a bingo.如果这些特定单元格的值设置为“true”,我会得到一个宾果游戏。 Otherwise, I continue to play the game.否则,我继续玩游戏。 I was looking for help to improve, or get feedback to completely re-haul this method for checking for a Bingo.我一直在寻求帮助以改进,或获得反馈以完全重新使用此方法来检查宾果游戏。

public int checkForBingo(int rowID, int colID) 
{    
    bool winner = false;

    for (int i = 0; i < 4; i++)
    {
        //checking verts and horiz
        if (winner = (boardState[i, 0]) == true && (boardState[i, 1] == true)
            && (boardState[i, 2] == true) && (boardState[i, 3] == true)
              && (boardState[i, 4] == true))
            break;
        else if (winner = (boardState[0, i] == true) && (boardState[1, i] == true)
            && (boardState[2, i] == true) && (boardState[3, i] == true) &&
            (boardState[4, i] == true))
            break;

        //checking diagonals
        if (!winner)
        {
            if (boardState[2, 2] == true)
                if ((winner = (boardState[0, 0] == true) && (boardState[1, 1] == true) &&
                    (boardState[3, 3] == true) && (boardState[4, 4] == true)))
                    break;
                else if (!(winner = (boardState[0, 4] == true) && (boardState[1, 3] == true) &&
                        (boardState[3, 1] == true) && (boardState[4, 0] == true)))
                    break;
        }
    }
    return 0;
}

For loop run from i=0 to 3 That means it's checking 4 rows only and in if condition you are checking for all 5 columns which is right. For 循环从 i=0 运行到 3 这意味着它只检查 4 行,并且在 if 条件下您正在检查所有 5 列是正确的。 Change your for loop condition to将您的 for 循环条件更改为

i <= 4

A few thing to note:有几点需要注意:

  • The main for loop should go be for (int i = 0; i < 5; i++) less than 5 ( arrays are 0 based)主要的 for 循环应该是for (int i = 0; i < 5; i++)小于 5(数组是基于 0 的)
  • You dont need to check for == true, a boolean value can be simple encapsulated in a if statement such as:您不需要检查 == true,布尔值可以简单地封装在 if 语句中,例如:

    if (boardState[0, 4] && boardState[1, 3] && boardState[3, 1] && boardState[4, 0]) if (boardState[0, 4] && boardState[1, 3] && boardState[3, 1] && boardState[4, 0])

  • The assignment in the if could lead to unwanted behavior(side effects) easy to forget is there. if的赋值可能会导致容易忘记的不良行为(副作用)。
  • The method signature should be capitalized ergo CheckForBingo方法签名应大写 ergo CheckForBingo
  • If you are not using the parameters - then remove them from the method signature如果您不使用参数 - 然后从方法签名中删除它们
  • As noted the code is a bit hard to follow如前所述,代码有点难以理解

For comparison and learning purpose(myself included) I have written the method like this:为了比较和学习的目的(包括我自己),我写了这样的方法:

public bool CheckForBingo(bool[,] boardState)
        {
            bool haveWeWon = false;

            //check horizotal
            for (int i = 0; i < 5; i++)
            {
                haveWeWon = true;

                for (int y = 0; y < 5; y++)
                {
                    if (!boardState[i, y])
                    {
                        haveWeWon = false;
                        break;
                    }
                }

                if(haveWeWon)
                {
                    return haveWeWon;
                }
            }

            if (!haveWeWon)
            {
                //check vertical
                for (int i = 0; i < 5; i++)
                {
                    haveWeWon = true;

                    for (int y = 0; y < 5; y++)
                    {
                        if (boardState[y, i])
                        {
                            haveWeWon = false;
                            break;
                        }
                    }

                    if (haveWeWon)
                    {
                        return haveWeWon;
                    }
                }
            }

            //check the middle - if false dont bother checking diagonal
            if (boardState[2, 2])
            {
                if (!haveWeWon)
                {
                    //check top left diagonal 
                    for (int i = 0; i < 5; i++)
                    {
                        haveWeWon = true;

                        if (!boardState[i, i])
                        {
                            haveWeWon = false;
                            break;
                        }
                    }

                    if (haveWeWon)
                    {
                        return haveWeWon;
                    }
                }

                if (!haveWeWon)
                {
                    //check top right diagonal 
                    for (int i = 4; i >= 0; i--)
                    {
                        haveWeWon = true;

                        if (!boardState[i, i])
                        {
                            haveWeWon = false;
                            break;
                        }
                    }

                    if (haveWeWon)
                    {
                        return haveWeWon;
                    }
                }
            }
            return false;
        }

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM