找到正方形2D陣列中任意一個隨機點的對角線

Question

我必須編寫設置8x8矩陣棋盤的代碼，然後詢問用戶他們想要放置皇后的行和列。然後我必須在女王可以移動的每個方格上放一個*。女王可以移動的行和列上的*號並不難，但我無法正確標記女王可以移動的對角線。這是迄今爲止我已經寫了嘗試並找到對角線代碼：

char[][] chessboard = new char[8][8]; 
      System.out.print("What row do you want to place the queen on? "); 
      int row = console.nextInt(); 
      System.out.print("What column do you want to place the queen on? "); 
      int column = console.nextInt(); 

      char queen = 'Q'; 
      chessboard[row - 1][column - 1] = queen; 

      // column and rows 
      for (int i = 0; i < chessboard.length; i++) { 
       for (int j = 0; j < chessboard[0].length; j++) { 
        if ((i == 2 || j == 6) && chessboard[i][j] != queen) { 
         chessboard[i][j] = '*'; 
        } 
       } 
      } 

      if ((row - 1) != 0) { 
       // left diagonal 
       for (int i = 0; i < row; i++) { 
        for (int j = (column - 1 - (row - 1)); ((column - 1) - (row - 1)) <= j && j < column; j++) { 
         if (chessboard[i][j] != queen) { 
          chessboard[i][j] = '*'; 
         } 
        } 
       } 
       for (int i = (row - 1) + (8 - (column)); i >= row - 1; i--) { 
        for (int j = 7; j >= (column - 1); j--) { 
         if (chessboard[i][j] != queen) { 
          chessboard[i][j] = '*'; 
         } 
        } 
       } 

       // right diagonal 
       for (int i = 7; i >= row - 1; i--) { 
        for (int j = (column - 1 - (row - 1)); 0 <= j && j < column; j--) { 
         if (chessboard[i][j] != queen) { 
          chessboard[i][j] = '*'; 
         } 
        } 
       } 

      } 
for (int i = 0; i < chessboard.length; i++) { 
      for (int j = 0; j < chessboard[0].length; j++) { 
       System.out.print(chessboard[i][j] + " "); 
      } 
      System.out.println(); 
     } 
    } 

當我把實驗值，例如第3行和列7，我得到一個非常混亂的輸出。對於上面的數字，我得到：

[][][][] * * * [] 
[][][][] * * * [] 
* * * * * * Q * 
* * * * * [] * * 
* * * * * [] * [] 
* * * * * [] * [] 
* * * * * [] * [] 
* * * * * [] * [] 

有人能告訴我我哪裏出錯了嗎？ *這是一個家庭作業問題，所以請只在答案中提供代碼。謝謝！

Answer 1

如果是我，我會簡單的通過棋盤的每個方格循環並測試每平方米的有效性作爲一個女王可以移動到，三個可能的規則，即

square.x == queen.x || 
square.y == queen.y || 
|square.x - queen.x| == |square.y - queen.y| 

如果你的廣場符合上述任何規則，那麼它是一個有效的廣場，否則它不是。省略廣場，女王當前所在，當然

square.x != queen.x && square.y != queen.y 

僞代碼：

for (int i = 0; i < chessboard.length; i++) { 
    for (int j = 0; j < chessboard[0].length; j++) { 
     // You could encapsulate these lines in a isValidSquareForQueenMove(x, y) method. 
     boolean isSquareValid = false; 
     isSquareValid = isSquareValid || x == queen.x; 
     isSquareValid = isSquareValid || y == queen.y; 
     isSquareValid = isSquareValid || Math.abs(queen.x - x) == Math.abs(queen.y - y); 
     isSquareValid = isSquareValid && x != queen.x && y != queen.y; 

     // Do stuff here. 
    } 
}