Question

我有一个递归函数，一个数组作为参数，当我从一个网格行进时存储路径，从（0,0）到（x，y），我必须跳过一些定义为“不可用”的点“

我实现了这样的功能

unAvailablePoint = [(1, 2), (3, 0), (0, 3), (2, 3), (0, 1)]

def steppable(point):
    return point not in unAvailablePoint

def travel(x, y, path, visited):
    if x >= 0 and y >= 0 and steppable((x, y)):
        if (x, y) in visited:
            return visited[(x, y)]
        success = False
        if (x, y) == (0, 0) or travel(x-1, y, path, visited) or travel(x, y-1, path, visited):
            path = path + [(x, y)] #the path will remain empty even after the recursive call have done some changes to the path
            success = True
        visited[(x, y)] = success
        return success
    return False

path = []
visited = {}
travel(3, 3, path, visited)
print(path) //[]

当我打印出最后的path时，似乎path仍为空。这不是我期望的Python新手。任何建议都会有所帮助

Answer 1

尝试追加到路径，并且不会在每次递归迭代时初始化它：

path.append( (x,y) ) #the path will remain empty even after the recursive call have done some changes to the path

而不是：

path = path + [(x, y)] #the path will remain empty even after the recursive call have done some changes to the path

这样，您不会在每次迭代时初始化列表，因此它不会是函数的局部变量。

Python - 理解变量传递到递归函数的范围

1 个答案: