请帮助我完成康威生命游戏的基本 java 实现

Question

我花了相当长的时间尝试编写一个程序来实现康威的生命游戏 - 包含更多信息的链接。 。我正在遵循一些在线指南，并获得了大部分功能。我编写了如下所示的“next”和“neighbours”方法。谁能告诉我这些是否是好的实现，以及如何做得更好？

练习的重点是不要修改或更改任何其他方法，而只编写下一个方法！ :)

import java.io.*;
import java.util.Random;

public class Life {

private boolean[][] cells;

public static void main( String[] args ) {
  Life generation = new Life( );
  for (int i = 0; i != 10; i++) {
    System.out.println( generation );
    generation.next( );
  }
}
// Constructors

public void next (){

  int SIZE;
  SIZE=cells.length;
  boolean[][] tempCells = new boolean [SIZE] [SIZE]; 

  for( int i=0; i<SIZE; i++ ) {
 for( int j=0; j<SIZE; j++ ) {
  tempCells[i][j] = cells[i][j];
 }
  } 
  for (int row = 0; row < cells.length ; row++)
  {
    for (int col = 0 ; col < cells[row].length ; col++)
    {
      if ( neighbours(row, col) > 3  ||  neighbours(row, col) < 2 )
      {
        tempCells[row][col] = false;
      }
      else if (neighbours(row, col) == 3 )
      {
        tempCells[row][col] = true;
      }      

    }

  }
  cells = tempCells;

}


public int neighbours (int row, int col) {
  int acc=0;
  for ( int i = row -1; i <= row + 1 ; i++)
    {
     for (int j = col -1 ; j <= col + 1 ; j++)
       {
       try {
         if (cells[i][j]==true && (i != row || j!=col))
         {
           acc++;
         }          
       } catch ( ArrayIndexOutOfBoundsException f)
       {continue;}
     }
  }
  return acc;
}


// Initialises 6 * 6 grid with Glider pattern.
public Life( ) {
final int SIZE = 8;
// Arguably, this should have been a class (static) array.
final int[][] pairs = {{2,4},{3,3},{1,2},{2,2},{3,2}};
cells = new boolean[ SIZE ][ ];
for (int row = 0; row < SIZE; row ++) {
cells[ row ] = new boolean[ SIZE ];
}
for (int pair = 0; pair < pairs.length; pair ++) {
final int row = pairs[ pair ][ 0 ];
final int col = pairs[ pair ][ 1 ];
cells[ row ][ col ] = true;
}
}
 // Initialise size * size grid with random cells.
//public Life( int size ) {
//final Random rand = new Random( );
//cells = new boolean[ size ][ ];
//for (int row = 0; row < size; row ++) {
//cells[ row ] = new boolean[ size ];
//for (int col = 0; col < size; col ++) {
//cells[ row ][ col ] = (rand.nextInt( 2 ) == 0);
//}
//}
//}
// Public methods and helper methods.

@Override
public String toString( ) {
String result = "";
for (int row = 0; row < cells.length; row ++) {
final boolean[] column = cells[ row ];
for (int col = 0; col < column.length; col ++) {
result = result + (column[ col ] ? "x" : ".");
}
result = result + "\n";
}
return result;
}
}

Answer 1

您无需将 cells 的内容复制到 tempCells（next 中的第一个嵌套循环）。相反，您可以在下一个循环中的 if-else 中添加一个额外的子句。此外，存储来自邻居的结果对于速度和清晰度来说可能是一个好主意。

for (int row = 0; row < cells.length ; row++)
    for (int col = 0 ; col < cells[row].length ; col++) {
       int n = neighbours(row,col);

       if (n > 3  ||  n < 2)
           tempCells[row][col] = false;
       else if (n == 3)
           tempCells[row][col] = true;
       else
           tempCells[row][col] = cells[row][col];
    }

（除此之外，看起来不错，但我还没有运行和测试你的代码。）

Answer 2

不要使用 ArrayIndexOutOfBoundException 来计算越界 (OOB) 条件。它会降低性能。最好使用环绕机制将数组视为球体，这样您就根本不会遇到 OOB。您可以尝试这样的操作：

public Cell[] getNeighbours(int i, int j) {
int i2 = i - 1, i3 = i + 1, j2 = j - 1, j3 = j + 1;
if (i2 == -1) i2 = board.length - 1;
if (i3 == (board.length)) i3 = 0;
if (j2 == -1) j2 = board[i].length - 1;
if (j3 == (board[i].length)) j3 = 0;
return new Cell[]{board[i2][j2], board[i2][j], board[i2][j3], board[i][j2], board[i][j3], board[i3][j2], board[i3][j], board[i3][j3]};

}

然后您可以循环返回的数组并检查其中有多少是活动的并返回该计数。