Question

我正在制作一个2048克隆，但它拒绝合并不在第一行/列交互的块，经过进一步调查后我发现for循环不是正确地完成工作

用于演示目的的示例代码：

import java.util.Arrays;

public class teste{
  public static void main(String[] args){
    int[] gameBoard = new int[] {4,2,2,0} ;

    gameBoard = pushLeft(gameBoard);
    System.out.println(Arrays.toString(gameBoard));

  }


  public static int[] pushLeft(int[] gameBoard) {
    System.out.println("Pushing left...");
    // This one is the troublemaker
    for (int y = 1; y < 4; y++) {
      System.out.println("Y = " + y);
      boolean[] alreadyCombined = { false, false, false, false };
      if(gameBoard[y] != 0) {
        int value = gameBoard[y];
        int aux = y-1;
        while ( (aux >= 0) && (gameBoard[aux] == 0)) {
          aux--;
        }
        if (aux == -1) {
          gameBoard[y] = 0;
          gameBoard[0] = value;
        }
        else if (gameBoard[aux] != value && y-aux != 1) {
          gameBoard[y] = 0;
          gameBoard[aux+1] = value;
        }
        else if(gameBoard[aux] != value) {
          break;
        }
        else {
          if(alreadyCombined[aux]) {
            gameBoard[y] = 0;
            gameBoard[aux+1] = value;
          }
          else {
            gameBoard[y] = 0;
            gameBoard[aux] *= 2;
            alreadyCombined[aux] = true;
          }
        }
      }
    }
    return gameBoard;
  }

此代码返回此输出：

Pushing left...
Y = 1
[4, 2, 2, 0]

正如你所看到的那样，它缺少Y = 2和Y = 3，这使得2个内部的两个人不能合并。有什么想法吗？

Answer 1

您的代码通过

else if(gameBoard[aux] != value) {
          break;
        }

因此，因为value = 2 et和gameBoard[aux] = 4所以中断会阻止您停止循环您应该尝试打印这些值以查看问题所在

Answer 2

代码

 else if(gameBoard[aux] != value) {
              break;
            }

在第一次迭代gabeBoard[aux] (4)中，而不是value (2)。所以它正在经历休息。

Answer 3

这个怎么样？它不是你要求的，而是你想要的更多： - ）

public static void main(String[] args) {
    int[] g = new int[] {4,2,2,0} ;

    pushLeft(g);
    System.out.println(Arrays.toString(g));
    g = new int[] {0,0,2,2} ;
    pushLeft(g);
    System.out.println(Arrays.toString(g));
    g = new int[] {4,2,0,0} ;
    pushLeft(g);
    System.out.println(Arrays.toString(g));
    g = new int[] {0,0,0,0} ;
    pushLeft(g);
    System.out.println(Arrays.toString(g));
    g = new int[] {0,0,0,2} ;
    pushLeft(g);
    System.out.println(Arrays.toString(g));

}

private static void pushLeft(int[] g) {
    for (int i = 0; i < g.length; i++) {
        // shift left
        for (int j = i+1; j < g.length; j++) {
            if (g[i] == 0 && g[j] > 0) {
                g[i] = g[j];
                g[j] = 0;
                break;
            }
        }
        // combine left
        for (int j = i+1; j < g.length; j++) {
            if (g[i] > 0 && g[i] == g[j]) {
                g[i] *= 2;
                g[j] = 0;
                break;
            }
        }
    }
}

打印：

[4, 4, 0, 0]
[4, 0, 0, 0]
[4, 2, 0, 0]
[0, 0, 0, 0]
[2, 0, 0, 0]

因为循环没有理由中断

3 个答案: