Question

我试图实现一个有序的链表。这是代码：

class Node:
    def __init__(self, initial_data):
        self.data = initial_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

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


    def __str__(self):
        output_string = ''

        current = self.head
        while current is not None:
            output_string += str(current.get_data())
            next_node = current.get_next()
            #gives the pointer to the next node i.e. the next node is that which is next to the current

            if next_node is not None:
                output_string += "->"

            current = next_node

        return output_string
    #does not need to be changed for ordered linked list
    def is_empty(self):
        if self.head is None:
            return True
        else:
            return False
    def insert(self, data):
        current = self.head
        previous = None
        stop = False
        while current is not None and not stop:
            if current.get_data() > data:
                stop = True
            else:
                previous = current
                current = current.get_next()
        temp = Node(data)
        if previous == None:
            temp.set_next(self.head)
            self.head = temp
        else:
            temp.set_next(current)
            previous.set_next(temp)
      def size(self):
        current = self.head
        count = 0
        while current != None:
            count += 1
            current = current.get_next()
        return count
    def search(self, item):
        current = self.head
        found = False
        stop = False
        while current is not None and not found and not stop:
            if current.get_data()== item:
                found = True
            else:
                if current.get_data() > item:
                    stop = True
                else:
                    current = current.get_next()
        return found
    def delete(self, item):
        current = self.head
        previous = None
        found = False
        while current is not None and not found:
            if current.get_data() == item:
                found = True
                break
            else:
                previous = current
                current = current.get_next()
        if found and previous is not None:
            previous.set_next(current.get_next())
        elif found and previous is None:
            self.head = None

以及测试用例：

def test_delete_smallest():
    my_list = LinkedList()
    my_list.insert(31)
    my_list.insert(77)
    my_list.insert(17)
    my_list.insert(93)
    my_list.insert(26)
    my_list.insert(54)

assert my_list.size() == 6
my_list.delete(17)
assert my_list.size() == 5

我很难过为什么它不会工作。

Answer 1

删除第一个元素时，您正在清除整个链接列表：

elif found and previous is None:
    self.head = None

此处previous为None，因为您从未将previous设置为其他任何内容。第一个元素current匹配。

您需要以不同方式处理此案例：

elif found and previous is None:
    self.head = current and current.get_next()

Python中的有序链接列表：AssertionError

1 个答案: