我正在尝试将链表按降序排序

Question

请问你能帮我吗？我正在尝试按照文档中的说明来排序链表，但我遇到了一个错误，错误信息是：
f.add('b', 2)
文件 "", 第 69 行，在 add 函数中
AttributeError: 'NoneType' 对象没有 'next' 属性。
我该如何避免这个问题呢？谢谢你。

class Frequency(object):
    """

    Stores a letter:frequency pair.

    >>> f = Frequency('c', 2)
    >>> f.letter
    'c'
    >>> f.frequency
    2
    >>> f
    {c: 2}
    """
    def __init__(self, letter, frequency):
        self.letter = letter
        self.frequency = frequency
        self.next = None

    def __repr__(self):
        return '{%s: %d}' % (self.letter, self.frequency)

class SortedFrequencyList(object):
    """
    Stores a collection of Frequency objects as a sorted linked list.
    Items are sorted from the highest frequency to the lowest.
    """
    def __init__(self):
        self.head = None

    def add(self, letter, frequency):
        """
        Adds the given `letter`:`frequency` combination as a Frequency object
        to the list. If the given `letter` is already in the list, the given
        `frequency` is added to its frequency.

        >>> f = SortedFrequencyList()
        >>> f.add('a', 3)
        >>> f
        ({a: 3})
        >>> f.add('b', 2)
        >>> f
            ({a: 3}, {b: 2})
        >>> f.add('c', 4)
        >>> f
        ({c: 4}, {a: 3}, {b: 2})
        >>> f.add('b', 3)
        >>> f
        ({b: 5}, {c: 4}, {a: 3})
        """

        current = self.head
        found = False
        if self.head is None:
            self.head = Frequency(letter, frequency)
        else:
            prev = None
            while current is not None:
                if current.letter == letter:
                    current.frequency = current.frequency + frequency
                    found = True
                prev = current
                current = current.next

                next1 = current.next
                if next1 is None:
                    current = next1

                if current.frequency < next1.frequency:
                    temp = current
                    current = next1
                    next1 = temp
                else:
                    current = next1
                    next1 = current.next.next


            if found is False:
                prev.next = Frequency(letter, frequency)

Answer 1

在这些代码行中

current = current.next
next1 = current.next

如果 current.next == None 会发生什么呢？

---

我不确定你是在做Python练习，还是因为你真的需要这个功能；如果是后者，其实Python里已经有一个内置的类可以帮你完成这个功能。这个类叫做 collections.Counter（在Python 2.7或3.x中都可以用）；如果你用的是更早的版本，可以通过继承 collections.defaultdict 自己创建一个。这个类也使用Python字典，而不是将数据存储为键值对。

举个例子：

>>> from collections import Counter
>>> x = Counter()
>>> x['a'] += 2
>>> x['b'] += 3
>>> x['c'] += 1
>>> x
Counter({'b': 3, 'a': 2, 'c': 1})

你可以用下面的代码恢复你数据的排序键值对表示：

x.most_common()