递归返回错误的列表列表

Question

我试图解决一个问题，其中一部分是找到从 (0, 0) 到二维数组最右边的所有路径。这是我的代码：

def route_finder_helper(x, y, current_path, filler, list_of_lists):

    current_path[filler] = (x, y)

    if x == 0 and y == 0:
        print(current_path)
        list_of_lists.append(current_path)
        return list_of_lists

    if x == 0:
        return route_finder_helper(x, y - 1, current_path, filler - 1, list_of_lists)

    if y == 0:
        return route_finder_helper(x - 1, y, current_path, filler - 1, list_of_lists)

    return route_finder_helper(x-1, y, current_path, filler - 1, list_of_lists) + \
           route_finder_helper(x, y-1, current_path, filler - 1, list_of_lists)

其中 x 和 y 是当前坐标，current_path 是当前路径的元组列表，filler 是要更改的列表位置的索引，list_of_lists 应该是所有路径。但是，当我运行这个程序并打印返回值时，我得到了这个输出：

 [(0, 0), (0, 1), (0, 2), (1, 2), (2, 2)]
 [(0, 0), (0, 1), (1, 1), (1, 2), (2, 2)]
 [(0, 0), (1, 0), (1, 1), (1, 2), (2, 2)]
 [(0, 0), (0, 1), (1, 1), (2, 1), (2, 2)]
 [(0, 0), (1, 0), (1, 1), (2, 1), (2, 2)]
 [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)]
 [[(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)]]

所以我得到了正确的路径，但我不知道如何将它们保存到列表列表中。有人可以帮我吗？

这就是我调用函数的方式：

x_coordinate = coordinates[0]  
y_coordinate = coordinates[1]  
path_length = (x_coordinate + 1) + (y_coordinate + 1) - 1  
start_filling = path_length - 1  
current_path = [0] * path_length  
paths = route_finder_helper(x_coordinate, y_coordinate, current_path, 
start_filling, [])

这是它应该返回的：

[[(0, 0), (0, 1), (0, 2), (1, 2), (2, 2)],[(0, 0), (0, 1), (1, 1), (1, 2), (2, 2)], [(0, 0), (1, 0), (1, 1), (1, 2), (2, 2)], [(0, 0), (0, 1), (1, 1), (2, 1), (2, 2)], [(0, 0), (1, 0), (1, 1), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)]]

Answer 1

这是正确答案：

def route_finder_helper(x, y, current_path, filler, list_of_lists):

    current_path[filler] = (x, y)

    if x == 0 and y == 0:
        return list_of_lists + [current_path[:]]

    if x == 0:
        return route_finder_helper(x, y - 1, current_path, filler - 1, list_of_lists)

    if y == 0:
        return route_finder_helper(x - 1, y, current_path, filler - 1, list_of_lists)

    return route_finder_helper(x-1, y, current_path, filler - 1, list_of_lists) + \
           route_finder_helper(x, y-1, current_path, filler - 1, list_of_lists)

x_coordinate = 2
y_coordinate = 2
path_length = (x_coordinate + 1) + (y_coordinate + 1) - 1
start_filling = path_length - 1
current_path = [0] * path_length
paths = route_finder_helper(x_coordinate, y_coordinate, current_path, start_filling, [])
print(paths)

输出：

[[(0, 0), (0, 1), (0, 2), (1, 2), (2, 2)], 
[(0, 0), (0, 1), (1, 1), (1, 2), (2, 2)], 
[(0, 0), (1, 0), (1, 1), (1, 2), (2, 2)], 
[(0, 0), (0, 1), (1, 1), (2, 1), (2, 2)], 
[(0, 0), (1, 0), (1, 1), (2, 1), (2, 2)], 
[(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)]]

这里的更正是：

return list_of_lists + [current_path[:]]

我使用current_path[:]添加一个副本

Answer 2

代码的逻辑是正确的，您遇到的问题是您正在处理一个可变项目的列表。这意味着每个引用列表的变量都会在您仅更改一个变量时获得更新的列表：

a = [1, 2, 3]
b = a
a[0] = -1
print(b)
>>> [-1, 2, 3]

当您附加到list_of_lists 时，您的代码中也会发生类似的情况。只需将附加更改为：

# return list_of_lists.append(current_path)  <- old code
return list_of_lists + [current_path]  <- new code

这保证了每次代码结束时您都会启动一个新列表，而不是附加到代码的其他部分也有引用的列表。

如果这对您来说是 Python 中的一个新概念，那么有很多不错的博客比我更详细地解释了这一点，例如here

完整代码：

def route_finder_helper(x, y, current_path, filler, list_of_lists):

    current_path[filler] = (x, y)

    if x == 0 and y == 0:
        return list_of_lists + [current_path]

    if x == 0:
        return route_finder_helper(x, y - 1, current_path, filler - 1, list_of_lists)

    if y == 0:
        return route_finder_helper(x - 1, y, current_path, filler - 1, list_of_lists)

    return route_finder_helper(x-1, y, current_path, filler - 1, list_of_lists) + \
           route_finder_helper(x, y-1, current_path, filler - 1, list_of_lists)

x_coordinate = 2
y_coordinate = 2
path_length = (x_coordinate + 1) + (y_coordinate + 1) - 1
start_filling = path_length - 1
current_path = [0] * path_length
paths = route_finder_helper(x_coordinate, y_coordinate, current_path, start_filling, [])
print(paths)
>>> [[(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)], [(0, 0), (1, 0), (2, 0), (2, 1), (2, 2)]]