Java 二维数组越界错误

Question

我正在编写一个处理二维数组的程序，根据周围的元素进行计算，然后创建一个新数组。首先，我创建了一个简单的 Java 程序来测试不同的情况。我想正确处理元素是否位于数组的边缘，如果是，则将 "up, down, left, or right" 设置为等于该元素。我的测试程序适用于位于顶部、左侧或不在边缘的元素，但不适用于底部或右侧。这是我当前的代码:

public class ArrayTest 
{ 
   public static void buildE(int[][] array, int y, int x)
   {
      int up; 
      int down; 
      int left;
      int right; 

      //if element is on the top left
      if (y == 0 && x == 0)
      {
         up = array[y][x];
         down = array[y + 1][x];
         left = array[y][x];
         right = array[y][x + 1];
      }

      //if element is on bottom right
      else if (y == array.length && x == array.length)
      {
         up = array[y - 1][x];
         down = array[y][x];
         left = array[y][x - 1];
         right = array[y][x];
      }

      //if element is on top right
      else if(y == 0 && x == array.length)
      {
         up = array[y][x];
         down = array[y + 1][x];
         left = array[y][x - 1];
         right = array[y][x];
      }

      //if element is on bottom left
      else if (y == array.length && x == 0)
      {
         up = array[y - 1][x];
         down = array[y][x];
         left = array[y][x];
         right = array[y][x + 1];
      }

      //if element is on top 
      else if (y == 0) 
      { 
         up = array[y][x];
         down = array[y + 1][x];
         left = array[y][x - 1];
         right = array[y][x + 1];
      }  

      //if element is on left
      else if (x == 0)
      {
         up = array[y - 1][x];
         down = array[y + 1][x];
         left = array[y][x];
         right = array[y][x + 1];
      }

      //if element is on bottom
      else if(y == array.length)
      {
         up = array[y - 1][x];
         down = array[y][x];
         left = array[y][x - 1];
         right = array[y][x + 1];
      }

      //if element is on right
      else if (x == array.length)
      {
         up = array[y - 1][x];
         down = array[y + 1][x];
         left = array[y][x - 1];
         right = array[y][x];
      }

      //if element is not on an edge 
      else
      {
         up = array[y - 1][x];
         down = array[y + 1][x];
         left = array[y][x - 1];
         right = array[y][x + 1];   
      }

      System.out.println();
      System.out.print("#####################################");
      System.out.println();
      System.out.println("Array Element: " + array[y][x]);
      System.out.println("Up: " + up);
      System.out.println("Down: " + down);
      System.out.println("Left: " + left);
      System.out.println("Right: " + right);
   }

   public static void outputArray(int[][] array)
   {
      for(int row = 0; row < array.length; row ++)
      {
         for (int column = 0; column < array[row].length; column++)
            System.out.printf("%d ", array[row][column]);
         System.out.println();
      }
   }

   public static void main(String[] args)
   {
      int [][] myArray = {{1, 12, 13, 14, 15}, {2, 22, 23, 24, 25},
         {3, 32, 33, 34, 35}, {4, 42, 43, 44, 45}, {5, 52, 53, 54, 55}};
      outputArray(myArray);
      buildE(myArray, 4, 0);
   }
}

此外，如果我设置 y = 4，它不会将其识别为 myArray.length。但是，如果我遍历直到 array.length，我认为这在我的实际程序中不会成为问题。任何帮助将非常感激!

Answer 1

数组索引从 0 开始，如果您的数组长度为 5，则最后一个元素将在第 4 个索引处访问。在你的代码中

  else if (y == array.length && x == array.length)
{

    up = array[y - 1][x];//Exception here
    down = array[y][x];
    left = array[y][x - 1];
    right = array[y][x];

}

将其更改为:

 else if (y == array.length-1 && x == array.length-1)
    {

        up = array[y - 1][x];
        down = array[y][x];
        left = array[y][x - 1];
        right = array[y][x];

    }

在其他 else if 中遵循相同的方法。