那些遇到过的问题

Java递归迷宫求解器问题

use*_*133 1 java recursion

我正在尝试使用递归编写一个迷宫求解器,它似乎尝试每个方向一次,然后停止,我无法弄清楚为什么.如果您发现问题,请告诉我.键0是开放空间1是墙2是路径3的一部分是迷宫的末端

public class Maze{
  public static void main(String[] args){
    int[][] maze = 
     {{1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1},
      {0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1},
      {1,0,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,1,1,0,1,0,1},
      {1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,1,0,1},
      {1,0,1,0,1,1,1,0,1,0,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,0,1},
      {1,0,1,0,1,0,1,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1},
      {1,0,1,1,1,0,1,0,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,0,1},
      {1,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,1,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,1},
      {1,1,1,0,1,1,1,1,1,1,1,1,1,0,1,0,1,0,1,0,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1},
      {1,0,1,0,1,0,0,0,0,0,0,0,1,0,1,0,1,0,1,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,1},
      {1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,0,1,1,1},
      {1,0,1,0,1,0,0,0,1,0,1,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1},
      {1,0,1,0,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1},
      {1,0,1,0,1,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1},
      {1,0,1,0,1,0,1,0,1,0,1,1,1,0,1,0,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,0,1,1,1,1,1,0,1},
      {1,0,0,0,1,0,1,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,1},
      {1,0,1,1,1,0,1,0,1,1,1,1,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,0,1},
      {1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,1,0,1},
      {1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,1,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1},
      {1,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,1},
      {1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,1,1,0,1,0,1,1,1,0,1,1,1,1,1,0,1,0,1,1,1,1,1,1,1,0,1,0,1,0,1,0,1},
      {1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,1,0,1,0,1,0,1},
      {1,1,1,0,1,0,1,1,1,0,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,0,1,0,1,1,1,0,1},
      {1,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,1},
      {1,0,1,1,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,0,1,0,1},
      {1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,1},
      {1,0,1,1,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,1,1,0,1,0,1,0,1},
      {1,0,0,0,0,0,0,0,1,0,1,0,1,0,1,0,0,0,1,0,1,0,1,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,1,0,1,0,0,0,1,0,1,0,1},
      {1,0,1,1,1,1,1,0,1,0,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,1,1,1,1,0,1},
      {1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,1},
      {1,1,1,1,1,0,1,1,1,1,1,1,1,0,1,0,1,0,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,0,1},
      {1,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,1},
      {1,0,1,1,1,0,1,1,1,0,1,0,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,0,1,0,1,0,1,0,1},
      {1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1},
      {1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1}};
    boolean[][] posCheck = new boolean[maze.length][maze[0].length];
    int r = 0;
    int c = 0;
    for(int row = 0; row < maze.length; row++){
      for(int col = 0; col < maze[row].length; col++){
        if(maze[row][col]==0){
          r = row;
          c = col;
        }
      }
    }
    maze[r][c] = 3;
    mazeSolver(1, 0, 0, maze, posCheck);
  }

  public static boolean mazeSolver(int r, int c, int d, int[][]maze, boolean[][] posCheck){
    maze[r][c] = 2;

    if(maze[r][c] == 3){
      print(maze);
      return true;
    }

    if((c+1 < maze.length) && maze[r][c+1]==0 && d != 1 && !posCheck[r][c+1]){
      if(d != 3)
        posCheck[r][c+1] = true;
      if(mazeSolver(r, c + 1, 3, maze, posCheck)){
        maze[r][c] = 2;
        return true;
      }
    }

    if((r-1 >= 0) && maze[r-1][c]==0 && !posCheck[r-1][c] && d != 2){
      if(d != 4)
        posCheck[r-1][c] = true;
      if(mazeSolver(r - 1, c, 4, maze, posCheck)){
        maze[r][c] = 2;
        return true;
      }
    }

    if((c-1 >= 0) && maze[r][c-1]==0 && !posCheck[r][c-1] && d != 3){
      if(d != 1)
        posCheck[r][c-1] = true;
      if(mazeSolver(r, c - 1, 1, maze, posCheck)){
        maze[r][c] = 2;
        return true;
      }
    }

    if((r+1 < maze.length) && maze[r+1][c]==0 && !posCheck[r+1][c] && d != 4){
      if(d != 2)
        posCheck[r+1][c] = true;
      if(mazeSolver(r + 1, c, 4, maze, posCheck)){
        maze[r][c] = 2;
        return true;
      }
    }

    print(maze);
    return false;
  }

  public static void print(int[][] maze){
    for(int row = 0; row<maze.length; row++){
      for(int col = 0; col<maze[row].length; col++)
        System.out.print(maze[row][col]);
      System.out.println();
    }
  }
}
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Bar*_*ney 5

看到你已经接受了答案,但无论如何我会添加它...

递归可以是解决某些问题的一种非常优雅的方法,但它可能需要一些时间才能解决问题.所以这不是一个确切的答案,为什么你的代码不起作用,但更多的是如何在这样的问题中使用递归的更高级别.

递归问题通常包含数据的两个部分:一些整体拼图状态,以及与当前尝试相关的一些状态.整个递归的工作原理是因为每次调用递归函数时都会将一些新状态推送到调用堆栈,当函数退出时,它会被移除,让您准备好尝试下一个选项.您还可以在递归函数中操纵整体拼图状态,但通常在您开始时我会建议您在函数中对拼图状态所做的任何更改都应在退出时恢复.

因此,在您的情况下,迷宫本身是拼图状态,当前路径是整个拼图状态的临时变化,并且当前位置是与当前调用堆栈相关联的瞬态.

所以整体解决方案开始采取以下形式:

  // global state
  private static int[][] maze;

  private static boolean solve(int r, int c) {
    // return true if I'm at the exit, false otherwise
  }
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主要功能只是提供起始坐标:

  public static void main(String[] args) {
    if (solve(1, 0)) {
      print();
    } else {
      System.out.println("no solution found");
    }
  }
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所以下一步是"解决"功能的主体(我在迷宫数据中将退出位置设置为3 - 请参见最后的完整解决方案),这将成为:

  private static boolean solve(int r, int c) {

    if (maze[r][c] == 3) {
      // we've found the exit
      return true;
    }

    // push the current position onto the path
    maze[r][c] == 2;

    // try up / down / left / right - if any of these return true then we're done
    if (available(r - 1, c) && solve(r - 1, c)) {
      return true;
    }
    if (available(r + 1, c) && solve(r + 1, c)) {
      return true;
    }
    if (available(r, c - 1) && solve(r, c - 1)) {
      return true;
    }
    if (available(r, c + 1) && solve(r, c + 1)) {
      return true;
    }

    // no result found from the current position so return false
    // ... but have to revert the temporary state before doing so
    maze[r][c] = 0;

    return false;
  }
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这首先检查我们是否在退出的简单情况,如果是这种情况则返回true.如果没有,我们将当前单元格推送到路径,并查找可用的邻居.如果我们找到一个,我们依次尝试每个,这是递归的核心...如果没有可用的邻居工作,那么我们失败了,并且必须回溯.

最后,如果我们回溯,我们必须从路径中删除当前单元格.

这就是它.'available'函数只检查潜在的单元格是否在边界而不是墙壁或已经在当前路径上:

  private static boolean available(int r, int c) {
    return r >= 0 && r < maze.length
        && c >= 0 && c < maze[r].length
        && (maze[r][c] == 0 || maze[r][c] == 3);
  }
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完整代码:

public class Maze2 {

  private static int[][] maze = 
     {{1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1},
      {0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1},
      {1,0,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,1,1,0,1,0,1},
      {1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,1,0,1},
      {1,0,1,0,1,1,1,0,1,0,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,0,1},
      {1,0,1,0,1,0,1,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1},
      {1,0,1,1,1,0,1,0,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,0,1},
      {1,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,1,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,1},
      {1,1,1,0,1,1,1,1,1,1,1,1,1,0,1,0,1,0,1,0,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1},
      {1,0,1,0,1,0,0,0,0,0,0,0,1,0,1,0,1,0,1,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,1},
      {1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,0,1,1,1},
      {1,0,1,0,1,0,0,0,1,0,1,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1},
      {1,0,1,0,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1},
      {1,0,1,0,1,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1},
      {1,0,1,0,1,0,1,0,1,0,1,1,1,0,1,0,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,0,1,1,1,1,1,0,1},
      {1,0,0,0,1,0,1,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,1},
      {1,0,1,1,1,0,1,0,1,1,1,1,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,0,1},
      {1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,1,0,1},
      {1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,1,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1},
      {1,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,1},
      {1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,1,1,0,1,0,1,1,1,0,1,1,1,1,1,0,1,0,1,1,1,1,1,1,1,0,1,0,1,0,1,0,1},
      {1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,1,0,1,0,1,0,1},
      {1,1,1,0,1,0,1,1,1,0,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,0,1,0,1,1,1,0,1},
      {1,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,1},
      {1,0,1,1,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,0,1,0,1},
      {1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,1},
      {1,0,1,1,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,1,1,0,1,0,1,0,1},
      {1,0,0,0,0,0,0,0,1,0,1,0,1,0,1,0,0,0,1,0,1,0,1,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,1,0,1,0,0,0,1,0,1,0,1},
      {1,0,1,1,1,1,1,0,1,0,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,1,1,1,1,0,1},
      {1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,1},
      {1,1,1,1,1,0,1,1,1,1,1,1,1,0,1,0,1,0,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,0,1},
      {1,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,1},
      {1,0,1,1,1,0,1,1,1,0,1,0,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,0,1,0,1,0,1,0,1},
      {1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1},
      {1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,3,1}};

  public static void main(String[] args) {
    if (solve(1, 0)) {
      print();
    } else {
      System.out.println("no solution found");
    }
  }

  private static boolean solve(int r, int c) {

    // if we're at the goal then we've solved it
    if (maze[r][c] == 3) {
      return true;
    }

    // mark the current cell as on the path
    maze[r][c] = 2;

    // try all available neighbours - if any of these return true then we're solved
    if (available(r - 1, c) && solve(r - 1, c)) {
      return true;
    }
    if (available(r + 1, c) && solve(r + 1, c)) {
      return true;
    }
    if (available(r, c - 1) && solve(r, c - 1)) {
      return true;
    }
    if (available(r, c + 1) && solve(r, c + 1)) {
      return true;
    }

    // nothing found so remove the current cell from the path and backtrack
    maze[r][c] = 0;

    return false;
  }

  // cell is available if it is in the maze and either a clear space or the
  // goal - it is not available if it is a wall or already on the current path
  private static boolean available(int r, int c) {
    return r >= 0 && r < maze.length
        && c >= 0 && c < maze[r].length
        && (maze[r][c] == 0 || maze[r][c] == 3);
  }

  // use symbols to make reading the output easier...
  private static final char[] SYMBOLS = {' ', '#', '.', '*' };

  private static void print(){
    for(int row = 0; row < maze.length; ++row) {
      for(int col = 0; col < maze[row].length; ++col) {
        System.out.print(SYMBOLS[maze[row][col]]);
      }
      System.out.println();
    }
  }
}
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最后,如果您想要打印出所有可能的解决方案而不是仅找到第一个解决方案,那么只需将解决后的函数顶部更改为:

// if we're at the goal then print it but return false to continue searching
if (maze[r][c] == 3) {
  print();
  return false;
}
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