メソッド内のJava多次元配列

Question

私は多次元配列を知っていましたが、基本的なことはよく知っていますが、それらをブールメソッドに結合することに固執しています。 （同じ場所に）別の配列へ：位置（列、行）とつのアレイを比較

• ：この場合、私は私のように機能しているこの方法を簡略化することができる方法を見つけるためにしようとしています。
• 両方の配列のすべての場所が一致する場合はtrueを返します。

私のコードには以下の方法がありますが、手作業で数字を「ハードコード」しない方法については適切な方法を知りたいのです。両方の配列を比較するためにネストされたforループを使用する必要がありますか？あなたは、ループについて学ぶために必要な支援のための

感謝〜

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

。

インデックスマッピング（2次元配列よりも扱いが簡単です）を使用して、1次元配列に2次元行列を格納することについてです。

しかし、この特定のケースでは：配列の長さに基づいて、ループのネスト

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

Answer 2

使用。

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; 
}