Question

给定一个二维char数组填充0＆＃39; s和1，其中0表示一个墙，1表示一个有效路径，我开发了一个名为findPath（int r，int c）的递归方法来查找出口标有“＆＃39; x”的迷宫。该方法接收迷宫的当前行和列并经过N，E，S，W方向，直到找到有效路径并用＆＃39; +＆＃39;标记该有效路径。如果发现所有方向被墙壁阻挡的情况，那么该方法将被假定为回溯直到不再是这种情况，然后用“F”标记该路径。象征着坏道路。

现在我无法弄清楚为什么findPath方法似乎并没有横穿所有方向，因为我的显示方法只显示程序从我传入的坐标开始而不是从那里移动到任何地方，为什么会这样呢？

这是我的Driver类

public class MazeMain2
{   
    public static void main(String[]args)
    {

    char[][] mazeArr = {{'0','0','0','1','0','0','0','0','0','0','0','0','0','0','0'},
                {'0','0','0','1','0','0','0','0','1','0','0','0','0','1','0'},
                {'0','0','0','1','1','1','1','1','1','1','1','1','0','0','0'},
                {'0','0','0','1','0','0','0','0','0','0','0','1','0','0','0'},
                {'0','0','0','1','1','1','1','1','0','0','0','1','0','0','0'},
                {'0','0','0','0','0','0','0','1','0','0','0','1','0','0','0'},
                {'0','0','0','0','1','1','1','1','0','0','0','1','0','0','0'},
                {'0','0','0','0','1','0','0','1','0','0','0','1','0','1','0'},
                {'0','0','0','0','1','0','0','1','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','0','1','1','1','1','1','1','1','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','0','1','0','0','0','0'},
                {'0','0','0','0','0','1','0','0','0','0','1','1','1','1','0'},
                {'0','0','0','0','0','0','0','0','0','0','1','0','0','0','0'}};

    MazeSolver2 mazeS = new MazeSolver2(mazeArr);

    mazeS.markEntry();
    mazeS.markExit();

    mazeS.solve(0, mazeS.start);


    }
}

这是我的迷宫求解器类和findPath方法

public class MazeSolver2
{
    int start;
    int exit;

    char[][] maze;

public MazeSolver2(char[][] currentMaze)
{
    maze = currentMaze;
}

//Finds where the first 1 is in the top row of the 
//maze (entrance)
public void markEntry()
{
    for(int x = 0; x < maze.length; x++)
    {
        if(maze[0][x] == '1')
        {
            maze[0][x] = 'E';
            start = x;
        }
    }
}

//Finds where the last 1 is in the bottom row of the 
//maze (exit)
public void markExit()
{
    for(int x = 0; x < maze.length; x++)
    {
        if(maze[maze.length - 1][x] == '1')
        {
            maze[maze.length - 1][x] = 'x';
            exit = x;
        }
    }
}

public void solve(int x, int y)
{
    if(findPath(x, y))
    {
        System.out.println(maze[x][y]);
    }
    else
        System.out.println("No solution");

}

public boolean findPath(int r, int c)
{   
    displayMaze(maze);

    //Found the exit
    if(maze[r][c] == 'x')
    {
        return true;
    }


    if(maze[r][c] == '0' || maze[r][c] == '+' || maze[r][c] == 'F')
    {
        return false;
    }

    maze[r][c] = '+';

    //If row is currently at zero then don't check north
    //direction because it will be outside of the maze
    if(r <= 0)
    {
        if(findPath(r, c++))
        {
            return true;
        }


        if(findPath(r++, c))
        {
            return true;
        }

        if(findPath(r, c--))
        {
            return true;
        }

    }

    else
    {
        //check N, E, S, W directions
        if(findPath(r--, c) || findPath(r, c++) ||
            findPath(r++, c) || findPath(r, c--))
        {
            return true;
        }
    }

    //Marking the bad path
    maze[r][c] = 'F';

    return false;

}

//Displays maze
public void displayMaze(char[][] maze)
{

        for(int row = 0; row < maze.length; row++)
        {
            for(int col = 0; col < maze.length; col++)
            {
                if(col == 14)
                {
                    System.out.print(maze[row][col]);
                    System.out.println();
                }
                else
                {
                    System.out.print(maze[row][col]);
                }
            }
        }   

    System.out.println();
    }
}

Answer 1

你的算法本身有几个流程，我不明白指出。您可以搜索迷宫遍历问题，并获得许多好的教程。

但是，请注意方法调用。请注意，如果使用findPath(int r, int c)调用findPath(5, 5)，则对findPath(r, c++)的调用会再次传递值findPath(5, 5)，而不会传递findPath(5, 6)。

因为在这种情况下findPath(r, c++)会被调用当前值c，然后c++被执行。

同样适用于findPath(r, c--) findPath(r++ , c)等等。

理解这一事实的一个好主意是在方法int r, int c的开头打印值findPath()。同时播放一些后增量/减量（x ++ / - x）和前增/减（++ x / - x）。

希望它有所帮助。

递归迷宫求解器方法问题

1 个答案: