如何简化在8拼图中移动拼贴的代码？

Question

在8谜题中，当它找到板子中的空白瓷砖（用0表示）时，它需要一次移动可以到达的所有相邻板子。

因为我在我的实现中将二维板映射到一维数组，所以在我的代码中使用index()确实有意义。

我无法弄清楚现在实施neighbors()的优雅方式，因此它现在涉及到相当多的冗余代码。

public class Board { 

private char[] tiles; 
private int N; 

private Board(char[] blocks) { 
    N = (int) Math.sqrt(blocks.length); 
    this.tiles = new char[blocks.length]; 
    System.arraycopy(blocks, 0, this.tiles, 0, N * N); 
} 

private void exch(int i, int j) { 
    char swap = tiles[i]; 
    tiles[i] = tiles[j]; 
    tiles[j] = swap; 
} 

public Iterable<Board> neighbors() 
{ 
    Stack<Board> neighbors = new Stack<>(); 
    for (int i = 0; i < N; i++) { 
     for (int j = 0; j < N; j++) { 
      if (tiles[index(i, j)] == 0) { 
       Board neighbor; 
       if (i > 0) { 
        neighbor = new Board(tiles); 
        neighbor.exch(index(i, j), index(i - 1, j)); 
        neighbors.push(neighbor); 
       } 

       if (j > 0) { 
        neighbor = new Board(tiles); 
        neighbor.exch(index(i, j), index(i, j - 1)); 
        neighbors.push(neighbor); 
       } 

       if (i < N - 1) { 
        neighbor = new Board(tiles); 
        neighbor.exch(index(i, j), index(i + 1, j)); 
        neighbors.push(neighbor); 
       } 

       if (j < N - 1) { 
        neighbor = new Board(tiles); 
        neighbor.exch(index(i, j), index(i, j + 1)); 
        neighbors.push(neighbor); 
       } 
       break; 
      } 
     } 
    } 
    return neighbors; 
} 
} 

Answer 1

public Stack<Board> update (int check, int checkB, int i, int, j, int index, int indexB, Stack<Board> neighbors, char[] tiles){ 
    if (check > checkB) { 
    neighbor = new Board(tiles); 
    neighbor.exch(index(i, j), index(i + index, j + indexB)); 
    neighbors.push(neighbor); 
    } 
    return neighbors; 
} 

这样称呼它，4次，每次条件:)

neighbor = update (i, 0, i, j, -1, 0, neighbor, tiles) 

将相当于：

if (i > 0) { 
    neighbor = new Board(tiles); 
    neighbor.exch(index(i, j), index(i - 1, j)); 
    neighbors.push(neighbor); 
} 

如果你想以减少变数功能，你也可以这样做：

public Stack<Board> update (boolean check, int i, int, j, int index, int indexB, Stack<Board> neighbor, char[] tiles){ 
    if (check) { 

然后只要把条件呼叫

neighbor = update ((i > 0), i, j, -1, 0, neighbor, tiles)