python 链表问题

发布网友发布时间：2022-05-25 17:33

共1个回答

热心网友时间：2023-11-25 05:10

class UnorderedList:
    def __init__(self):
        self.head = None

    def add(self, item):
        new_node = Node(item)
        new_node.set_next(self.head)
        self.head = new_node

    def size(self):
        current = self.head
        count = 0
        while current != None:
            count = count + 1
            current = current.get_next()
        return count

    def is_empty(self):
        return self.head == None

    def add_all(self, other_list):
        for item in other_list:
            self.add(item)
        return self.head

    def __iter__(self):
        current = self.head
        while current:
            yield current
            current = current.get_next()

    def print_all(self):
        for item in self:
            print(item.get_data())

class Node:
    def __init__(self, init_data):
        self.data = init_data
        self.next = None

    def get_data(self):
        return self.data

    def get_next(self):
        return self.next

    def set_data(self, new_data):
        self.data = new_data

    def set_next(self, new_next):
        self.next = new_next

if __name__ == "__main__":
    s = UnorderedList()
    s.add(33)
    s.add(44)
    s.add_all([1,2,3,4])
    s.print_all()

追问您好，请问需不需要强调一下while current ！= None。

追答class UnorderedList:
    def __init__(self):
        self.head = None

    def add(self, item):
        self.head = Node(item).set_next(self.head)

    def size(self):
        current = self.head
        count = 0
        while current:
            count = count + 1
            current = current.get_next()
        return count
    ...

class Node:
    ...
    def set_next(self, new_next):
        self.next = new_next
        return self