java - 递归迷宫求解器方法问题

给定一个用0和1填充的二维char数组，其中0表示墙，1表示有效路径，我开发了一种名为findPath（int r，int c）的递归方法，以在迷宫中找到标记为'x的出口'。该方法进入迷宫的当前行和列，并经过N，E，S，W方向，直到找到有效路径并用“ +”标记该有效路径。给定一个实例，发现所有方向都被墙堵住了，那么该方法应该回溯直到不再是这种情况，然后用“ F”标记该路径，以表示该路径不正确。

现在我无法弄清楚为什么findPath方法似乎并不能横向遍历所有方向，因为我的显示方法只是显示了从我传入的坐标开始而不是从那里移动的程序，为什么会这样呢？

这是我的司机课

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);


    }
}

这是我的带有findPath方法的迷宫求解器类

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();
    }
}

最佳答案

您的算法本身具有多个流程，我不认为这是正确的。您可以搜索迷宫遍历问题，并获得许多不错的教程。

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++)的调用将再次传递值findPath(5, 5)，而不是使用findPath(5, 6)。

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

findPath(r, c--) findPath(r++ , c)等也是如此。

了解事实的一个好主意是在方法int r, int c的开头打印值findPath()。
也可以使用后期增量/减量（x ++ /-x）和前置增量/减量（++ x /-x）玩一点。

希望能帮助到你。