18x18板陣列索引出界

Question

// Counts the neighbors of alive or dead cells in boolean grid. 
    public static int countNeighbors (final boolean[][] grid, final int row, final int col) { 
     // Finds neighbors in top row. 
     int count = 0; 
     for (int i = 1; i > -1; --i) { 
      if (grid[row - 1][col + i] == true) 
       count += 1; 
      else if (grid[row - 1][col + i] == false) 
       count += 0; 
     } 

     // Finds neighbors in same row. 
     for (int i = 1; i > -1; --i) { 
      if (grid[row][col + i] == true) 
       count += 1; 
      else if (grid[row][col + i] == false) 
       count += 0; 
     } 

     // Finds neighbors in bottom row. 
     for (int i = 1; i > -1; --i) { 
      if (grid[row + 1][col + i] == true) 
       count += 1; 
      else if (grid[row + 1][col + i] == false) 
       count += 0; 
     } 

     return count; 
    } 

當我試圖在指定的正方形周圍的所有8個塊中查找所有真正的鄰居值時，獲取數組越界。

我覺得代碼已經可以處理，如果它超出了界限，因爲我認爲這些值將是錯誤的。

Answer 1

創建一個單獨的函數來獲取網格單元，包括邊界檢查：

public static boolean getGridCell (final boolean[][] grid, final int row, final int col) 
{ 
    // bounds check: 
    if((row < 0) || (row >= grid.length)) 
     return false; 
    if((col < 0) || (col >= grid[row].length)) 
     return false; 

    return grid[row][col]; 
} 

然後調用這個函數，而不是直接訪問網格：

public static int countNeighbors (final boolean[][] grid, final int row, final int col) { 
    // Finds neighbors in top row. 
    int count = 0; 
    for (int i = 1; i >= -1; --i) { 
     if (getGridCell(grid, row - 1,col + i)) 
      count += 1; 
    } 

    // Finds neighbors in same row. 
    for (int i = 1; i >= -1; --i) { 
     if (getGridCell(grid, row, col + i)) 
      count += 1; 
    } 

    // Finds neighbors in bottom row. 
    for (int i = 1; i >= -1; --i) { 
     if (getGridCell(grid, row + 1, col + i)) 
      count += 1; 
    } 

    return count; 
} 

另外：

• 沒有必要檢查網格單元是否爲空，並且如果它是零，則計數，因爲這不會產生任何差異對伯爵的看法。你不需要測試if（some_boolean == true），你可以直接寫if（some_boolean）
• 你的循環終止條件應該是> = -1 not> -1，如果你打算包括-1 。