方法中的Java多維數組

Question

我知道多維數組，並且有點基礎，但是我堅持將它們組合成布爾方法。 在這種情況下，我試圖找出如何簡化此方法的方法，該方法的作用是：

• 將一個具有位置（行：列）的數組與另一陣列（具有相同位置）進行比較。
• 如果兩個數組中的所有位置都匹配，則應返回true。

下面的方法適用於我的代碼，但是我想知道如何不單手“硬編碼”數字的正確方法。 我是否需要使用嵌套的for循環來比較兩個數組？

謝謝你的幫助〜

public static boolean gameIsWon(int[][] workingPuzzle, int[][] solvedPuzzle)
{
    if (workingPuzzle[0][0] == solvedPuzzle[0][0] &&
            workingPuzzle[0][1] == solvedPuzzle[0][1] &&
            workingPuzzle[0][2] == solvedPuzzle[0][2] &&
            workingPuzzle[0][3] == solvedPuzzle[0][3] &&
            workingPuzzle[0][4] == solvedPuzzle[0][4] &&
            workingPuzzle[0][5] == solvedPuzzle[0][5] &&
            workingPuzzle[0][6] == solvedPuzzle[0][6] &&
            workingPuzzle[0][7] == solvedPuzzle[0][7] &&
            workingPuzzle[0][8] == solvedPuzzle[0][8] &&
            workingPuzzle[1][0] == solvedPuzzle[1][0] &&
            workingPuzzle[1][1] == solvedPuzzle[1][1] &&
            workingPuzzle[1][2] == solvedPuzzle[1][2] &&
            workingPuzzle[1][3] == solvedPuzzle[1][3] &&
            workingPuzzle[1][4] == solvedPuzzle[1][4] &&
            workingPuzzle[1][5] == solvedPuzzle[1][5] &&
            workingPuzzle[1][6] == solvedPuzzle[1][6] &&
            workingPuzzle[1][7] == solvedPuzzle[1][7] &&
            workingPuzzle[1][8] == solvedPuzzle[1][8] &&
            workingPuzzle[2][0] == solvedPuzzle[2][0] &&
            workingPuzzle[2][1] == solvedPuzzle[2][1] &&
            workingPuzzle[2][2] == solvedPuzzle[2][2] &&
            workingPuzzle[2][3] == solvedPuzzle[2][3] &&
            workingPuzzle[2][4] == solvedPuzzle[2][4] &&
            workingPuzzle[2][5] == solvedPuzzle[2][5] &&
            workingPuzzle[2][6] == solvedPuzzle[2][6] &&
            workingPuzzle[2][7] == solvedPuzzle[2][7] &&
            workingPuzzle[2][8] == solvedPuzzle[2][8] &&
            workingPuzzle[3][0] == solvedPuzzle[3][0] &&
            workingPuzzle[3][1] == solvedPuzzle[3][1] &&
            workingPuzzle[3][2] == solvedPuzzle[3][2] &&
            workingPuzzle[3][3] == solvedPuzzle[3][3] &&
            workingPuzzle[3][4] == solvedPuzzle[3][4] &&
            workingPuzzle[3][5] == solvedPuzzle[3][5] &&
            workingPuzzle[3][6] == solvedPuzzle[3][6] &&
            workingPuzzle[3][7] == solvedPuzzle[3][7] &&
            workingPuzzle[3][8] == solvedPuzzle[3][8] &&
            workingPuzzle[4][0] == solvedPuzzle[4][0] &&
            workingPuzzle[4][1] == solvedPuzzle[4][1] &&
            workingPuzzle[4][2] == solvedPuzzle[4][2] &&
            workingPuzzle[4][3] == solvedPuzzle[4][3] &&
            workingPuzzle[4][4] == solvedPuzzle[4][4] &&
            workingPuzzle[4][5] == solvedPuzzle[4][5] &&
            workingPuzzle[4][6] == solvedPuzzle[4][6] &&
            workingPuzzle[4][7] == solvedPuzzle[4][7] &&
            workingPuzzle[4][8] == solvedPuzzle[4][8] &&
            workingPuzzle[5][0] == solvedPuzzle[5][0] &&
            workingPuzzle[5][1] == solvedPuzzle[5][1] &&
            workingPuzzle[5][2] == solvedPuzzle[5][2] &&
            workingPuzzle[5][3] == solvedPuzzle[5][3] &&
            workingPuzzle[5][4] == solvedPuzzle[5][4] &&
            workingPuzzle[5][5] == solvedPuzzle[5][5] &&
            workingPuzzle[5][6] == solvedPuzzle[5][6] &&
            workingPuzzle[5][7] == solvedPuzzle[5][7] &&
            workingPuzzle[5][8] == solvedPuzzle[5][8] &&
            workingPuzzle[6][0] == solvedPuzzle[6][0] &&
            workingPuzzle[6][1] == solvedPuzzle[6][1] &&
            workingPuzzle[6][2] == solvedPuzzle[6][2] &&
            workingPuzzle[6][3] == solvedPuzzle[6][3] &&
            workingPuzzle[6][4] == solvedPuzzle[6][4] &&
            workingPuzzle[6][5] == solvedPuzzle[6][5] &&
            workingPuzzle[6][6] == solvedPuzzle[6][6] &&
            workingPuzzle[6][7] == solvedPuzzle[6][7] &&
            workingPuzzle[6][8] == solvedPuzzle[6][8] &&
            workingPuzzle[7][0] == solvedPuzzle[7][0] &&
            workingPuzzle[7][1] == solvedPuzzle[7][1] &&
            workingPuzzle[7][2] == solvedPuzzle[7][2] &&
            workingPuzzle[7][3] == solvedPuzzle[7][3] &&
            workingPuzzle[7][4] == solvedPuzzle[7][4] &&
            workingPuzzle[7][5] == solvedPuzzle[7][5] &&
            workingPuzzle[7][6] == solvedPuzzle[7][6] &&
            workingPuzzle[7][7] == solvedPuzzle[7][7] &&
            workingPuzzle[7][8] == solvedPuzzle[7][8] &&
            workingPuzzle[8][0] == solvedPuzzle[8][0] &&
            workingPuzzle[8][1] == solvedPuzzle[8][1] &&
            workingPuzzle[8][2] == solvedPuzzle[8][2] &&
            workingPuzzle[8][3] == solvedPuzzle[8][3] &&
            workingPuzzle[8][4] == solvedPuzzle[8][4] &&
            workingPuzzle[8][5] == solvedPuzzle[8][5] &&
            workingPuzzle[8][6] == solvedPuzzle[8][6] &&
            workingPuzzle[8][7] == solvedPuzzle[8][7] &&
            workingPuzzle[8][8] == solvedPuzzle[8][8]
        )
        return true;
    else
        return false;
}

Answer 1

您需要了解循環。

關於使用索引映射將二維矩陣存儲在一維數組中（比二維數組更易於處理）。

但是在這種情況下：

return java.util.Arrays.deepEquals(workingPuzzle, solvedPuzzle);

Answer 2

根據數組長度使用嵌套循環。

 public static boolean gameIsWon(int[][] workingPuzzle, int[][] solvedPuzzle) {
  if(workingPuzzle.length != solvedPuzzle.length) {
      return false; 
  }
  for(int i = 0; i < workingPuzzle.length; i++) {
    if(workingPuzzle[i].length != solvedPuzzle[i].length) {
      return false;
    }
    for(int j = 0; j < workingPuzzle[i].length; j++) {
      if(workingPuzzle[i][j] != solvedPuzzle[i][j]) {
          return false; 
      }
    }
  }
  return true;
}