Question

我应该计算有多少@符号连接到2D阵列矩阵中的原始预定点。我收到堆栈溢出错误，不知道为什么。任何建议都表示赞赏（获取提供的行和列位置，并计算有多少@符号连接到原始位置。如果它们相互连接，向下，向左和向右连接，则连接@符号。）

当前代码：

   import static java.lang.System.*;

public class AtCounter
{
   private String[][] atMat;
   private int totalCount = 0;
   private boolean[][] visited; //used to see if location has been visited before.
   public AtCounter(int rows, int cols)
   {
       //size the matrix
       atMat = new String[rows][cols];
       //use nested loops to randomly load the matrix
       for(int r = 0; r < atMat.length; r++)
       {
           for(int c = 0; c < atMat[r].length; c++)
           {
               int num = (int) (Math.random() * 2);
               if(num == 0)
                   atMat[r][c] = "@";
               else
                   atMat[r][c] = "-";
           }
       }
       //you will need to use Math.random()

        visited = new boolean[atMat.length][atMat.length];
   }
   /**
    * Used to find out if location is in the 2D Matrix.
    */
   public boolean inBounds(int r, int c)
   {
        return ((r > -1 && r < atMat.length) && (c > -1 && c < atMat.length));
   }
   public int countAts(int r, int c)
   {
        //add in recursive code to count up the # of @s connected
        //start checking at spot [r,c]

        if(atMat[r][c].equals("-") || !inBounds(r,c))
            return 0;
        if(!visited[r][c])
        {
            if(atMat[r][c].equals("@"))
            {
                totalCount+=1;

                if(inBounds(r - 1, c))//up
                    countAts(r - 1, c);

                if(inBounds(r + 1, c))//down
                    countAts(r + 1, c);

                if(inBounds(r, c + 1))//right
                    countAts(r , c + 1);

                if(inBounds(r, c  - 1))//left
                    countAts(r, c - 1);
            }
        }
        return totalCount;        
     }


    /*
     *this method will return all values in the matrix
     *this method should return a view of the matrix
     *that looks like a matrix
     */
    public String toString()
    {
        String grid = "";
        for(String[] row : atMat)
        {
            for(String val : row)
            {
                grid += val + " ";
            }
            grid += "\n";
        }
        return grid;
    }
}

Answer 1

您的代码中有几处错误：

您使用visited创建atMat.length数组，即使您的原始尺寸（行和 cols ）不是，也会创建一个正方形数组等于。
在inBounds方法中，列参数c与行的长度有关，而不是列的长度
在countAts方法中：
- 无效条件的检查顺序需要反转，您需要首先检查它是否是有效位置，如果该单元格中的值是@，则需要先检查。
- 如果当前的单元格没有被访问过，那么第一件事是在if块中标记为被访问，以避免陷入无休止的递归。

尽管如此，可能的解决方案如下：

import static java.lang.System.*;

public class AtCounter
{
    private String[][] atMat;
    private int totalCount = 0;
    private boolean[][] visited; //used to see if location has been visited before.

    private int rows; // To store rows length
    private int cols; // To store cols length

    public AtCounter(int rows, int cols)
    {
        //size of the matrix
        this.rows = rows;
        this.cols = cols;
        atMat = new String[rows][cols];

        //use nested loops to randomly load the matrix
        for(int r = 0; r < rows; r++)
            for(int c = 0; c < cols; c++) {
                int num = (int) (Math.random() * 2);
                if(num == 0)
                    atMat[r][c] = "@";
                else
                    atMat[r][c] = "-";
            }

        visited = new boolean[rows][cols];
    }

    /**
    * Used to find out if location is in the 2D Matrix.
    */
    public boolean inBounds(int r, int c)
    {
        return r > -1 && r < rows && c > -1 && c < cols;
    }

    public int countAts(int r, int c)
    {
        //add in recursive code to count up the # of @s connected
        //start checking at spot [r,c]

        if(!inBounds(r,c) || atMat[r][c].equals("-")) // The order here matters
            return 0;

        if(!visited[r][c])
        {
            visited[r][c] = true; // Marks the current cell as visited
            if(atMat[r][c].equals("@"))
            {
                totalCount+=1;

                if(inBounds(r - 1, c))//up
                    countAts(r - 1, c);

                if(inBounds(r + 1, c))//down
                    countAts(r + 1, c);

                if(inBounds(r, c + 1))//right
                    countAts(r , c + 1);

                if(inBounds(r, c  - 1))//left
                    countAts(r, c - 1);
            }
        }
        return totalCount;
    }

}

使用递归计算Matrix Java中的连接字符串

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