通过迷宫的最短路径忽略一面墙

Question

所以我正在接受foo.bar挑战并需要准备兔子逃脱。每当我在我的代码上运行一个测试用例时它就可以工作，但仍然说提交失败了。

挑战在于获得通过矩阵迷宫m x n的最短路径，其中2 <1。 m，n＆lt; 20和一面墙可以被移除。

我的解决方案是从一开始就找到完成所有可能的路径，计算沿途穿过的墙数。如果墙的数量为1或更少，则解决方案有效并记录其长度。

path_dict = {}
min_solution = None

def answer(maze):
    path = None
    traverse (maze, path, 0, 0)
    global min_solution
    return min_solution

def traverse (maze, path, x, y):
    global path_dict
    global min_solution 

    cell_str = '(%d,%d)' % (x, y)

    x_max = len (maze) - 1
    y_max = len (maze[0]) - 1

    # check x and y are valid
    if x < 0 or x > x_max:
        return 
    if y < 0 or y > y_max:
        return 

    # Check not already visited cell
    if path and cell_str in path:
        return 

    total = None
    if path:
        total = path_dict [path]

    f_cell = '(%d,%d)' % (x_max, y_max)

    if path and f_cell in path:
        return

    # Check if final cell
    if x == x_max and y == y_max:
        if not path [-5:] == f_cell:
            path += f_cell
            path_dict [path] = total
        solution_len = len (path.split (')('))
        if not min_solution:
            min_solution = solution_len
        elif solution_len < min_solution:
            min_solution = solution_len
        #maze [x][y] = 2 
        return 

    #if path and path [-5:] == f_cell:
    #   return 
    # Not final cell - get value, add to count

    # Add path string to mem
    if path:
        path += cell_str
        cell_val = maze [x][y]
        total += cell_val
        # too many walls hit, stop checking this path
        if total >= 2:
            return 
    else:
        path = cell_str
        total = 0

    path_dict [path] = total

    if min_solution and len (path.split (')(')) > min_solution:
        return

    # Move DOWN, RIGHT, UP, LEFT
    traverse (maze, path, x + 1, y)
    traverse (maze, path, x, y + 1)
    traverse (maze, path, x - 1, y)
    traverse (maze, path, x, y - 1)

    return