使用递归返回列表中不包括 K 和以下数字的总和

Question

我的代码应该接收一个整数和一个整数列表。它返回 列表中的整数之和除了，它忽略 不吉利的数字和紧跟在它后面的数字 倒霉的号码。我只是尝试使用递归（无迭代）。到目前为止，我的代码适用于前 5 个案例，但后 2 个案例存在问题，因为它们最后有不吉利的数字。我不会试图完全重写我的函数，只是在我目前拥有的内容中发现错误。

def unlucky(unlucky_num, a_list):
    """
    >>> unlucky(13,[1])
    1
    >>> unlucky(13,[1,2,3,4,5,6])
    21
    >>> unlucky(13,[13,1,2,3])
    5
    >>> unlucky(13,[1,13,2,3])
    4
    >>> unlucky(13,[13, 0])
    0
    >>> unlucky(13,[13,1,2,13,2,1,13]) #13,1,13,2,13 ignored
    3
    >>> unlucky(7,[7,4,5,7,5,4,7]) #7,4,7,5,7 ignored (7 is unlucky)
    9
    """

    if a_list == []:
        return 0
    if a_list[0] == unlucky_num:
        if len(a_list) > 1:
            return unlucky(unlucky_num, a_list[2:])
        return unlucky(unlucky_num, a_list)
    return a_list[0] + unlucky(unlucky_num, a_list[1:]) 

对于最后两种情况，我收到以下错误（文件名已编辑）：

Error
**********************************************************************
File "------", line 39, in sum_unlucky
Failed example:
    sum_unlucky(13,[13,1,2,13,2,1,13]) #13,1,13,2,13 ignored
Exception raised:
    Traceback (most recent call last):
      File "-------", line 138, in __run
        exec(compile(example.source, filename, "single",
      File "<doctest sum_unlucky[5]>", line 1, in <module>
        sum_unlucky(13,[13,1,2,13,2,1,13]) #13,1,13,2,13 ignored
      File "-------", line 50, in sum_unlucky
        return sum_unlucky(unlucky_num, a_list[2:])
      File "---------", line 52, in sum_unlucky
        return a_list[0] + sum_unlucky(unlucky_num, a_list[1:])
      File "--------", line 50, in sum_unlucky
        return sum_unlucky(unlucky_num, a_list[2:])
      File "--------", line 52, in sum_unlucky
        return a_list[0] + sum_unlucky(unlucky_num, a_list[1:])
      File "-------", line 51, in sum_unlucky
        return sum_unlucky(unlucky_num, a_list)
      File "------", line 51, in sum_unlucky
        return sum_unlucky(unlucky_num, a_list)
      File "------", line 51, in sum_unlucky
        return sum_unlucky(unlucky_num, a_list)
      [Previous line repeated 986 more times]
      File "----", line 45, in sum_unlucky
        if a_list == []: 
    RecursionError: maximum recursion depth exceeded in comparison

Answer 1

您实际上只需要两个测试，这将更容易避免错误。对基本情况的测试，它是一个空数组和对倒霉数的测试。您不需要在基本情况之后测试长度，因为您可以传回一个空数组。这样，该函数就可以以一种非常易读的方式反映需求的方式编写：

def unlucky(unlucky_num, a_list):
    if not a_list:                             # base case
        return 0 

    head, *rest = a_list                       # rest will be [] when len(a_list) == 1

    if head == unlucky_num:
        return unlucky(unlucky_num, rest[1:])
    return head + unlucky(unlucky_num,rest)

unlucky(7,[7,4,5,7,5,4,7])
# 9

Answer 2

如果len(a_list) == 1，您忘记排除不吉利的数字。使用a_list[1:]。

def unlucky(unlucky_num, a_list):
    if a_list == []:
        return 0

    if a_list[0] == unlucky_num:
        if len(a_list) > 1:
            return unlucky(unlucky_num, a_list[2:])
        return unlucky(unlucky_num, a_list[1:])  # <- Here
    return a_list[0] + unlucky(unlucky_num, a_list[1:])

但在这种情况下，由于a_list[1:] == []，您可以跳过递归调用而只需return 0。你也可以简化：

def unlucky(unlucky_num, a_list):
    if a_list in ([], [unlucky_num]):
        return 0
    elif a_list[0] == unlucky_num:
        return unlucky(unlucky_num, a_list[2:])
    return a_list[0] + unlucky(unlucky_num, a_list[1:])

Answer 3

正如 Mark Meyer 所说，它永远不会到达 a_list == []。试试：

def unlucky(unlucky_num, a_list):
    if len(a_list) == 1:
        if a_list[0] == unlucky_num:
            return 0
        else:
            return a_list[0]
    if a_list[0] == unlucky_num:
        if len(a_list) > 1:
            return unlucky(unlucky_num, a_list[2:])
        return unlucky(unlucky_num, a_list)
    return a_list[0] + unlucky(unlucky_num, a_list[1:])