Python 3. 从链表中删除项目

Question

我在python中实现了链表，它完美地添加了元素并打印了列表。 但我对删除方法有问题。我想从列表中删除最后一项

这是我的班级 Cell()：

class Cell():
def __init__(self, value = None, next = None):
    self.value = value
    self.next = next

这是我的类 LinkedList()：

class LinkedList():
def __init__(self):
    self.top = None
    self.last = None
    self.length = 0

添加新单元的方法效果很好：

    def add (self, value, position = None):

    newCell = Cell(value, None)
    self.length += 1
    if position is None or position >= self.length:
        if self.top == None:
            self.last = self.top = Cell(value, None)
        else:
            self.last.next = self.last = Cell(value, None)
    elif position == 1:
        self.top = Cell (value, self.top)
    elif position > 1:
        afterMe = self.top
        i = 1
        for i in range(position-2):
            afterMe = afterMe.next
        newCell.next = afterMe.next
        afterMe.next = newCell

toString() 方法也很好用：

    def __str__(self):

    linkedList = ''
    cell = self.top
    if self.top is None:
        return 'Linked list is empty'

    for i in range(self.length):
        if cell == self.last:
            linkedList += (str(cell.value))
        else:
            linkedList += (str(cell.value)+ ', ')

        cell = cell.next
    return linkedList

这是我的删除方法，它会产生错误：

def delete(self, value = None):  # want to delete last cell
    if self.top == None:
        return None
    current = self.top

    if value is None:
        self.length -= 1

        while (current.next != self.last):
            current = current.next

        self.last = current
        current.next = None

    else:
        while (current.next.value != value):
            current = current.next
            if current == self.last:
                print ('no such value')
                return
        current.next = current.next.next

以下是代码的工作原理和错误：

numbers = LinkedList()
numbers.add(55)
numbers.add(75)
numbers.add(65)
print(numbers) # 55, 75, 65
numbers.add(3,2)
numbers.add (40,3)
print(numbers) # 55, 3, 40, 75, 65
numbers.delete()
print(numbers) # 55, 3, 40, 75
numbers.delete(40)
print(numbers)

    ''' returns error:
    Traceback (most recent call last):
    File "C:/Users/demin.va/Documents/Dropbox/Programming/Алгоритмы/связные списки.py", line 105, in <module>
    print(numbers)
  File "C:/Users/demin.va/Documents/Dropbox/Programming/Алгоритмы/связные списки.py", line 72, in __str__
    linkedList += (str(cell.value)+ ', ')
AttributeError: 'NoneType' object has no attribute 'value'
    '''

请回答，我应该如何更改代码以正确删除最后一个单元格或从不同位置删除？

Answer 1

current.next=None 应该在 while 循环之外：

def delete (self): #want to delete last cell
    if self.top == None:
        return None
    self.length -=1
    current = self.top
    nextcell = self.top.next
    while (nextcell != None):
        if nextcell == self.last:
            current = self.last
            break
        current = current.next
        nextcell = nextcell.next
    # current.next should be run only after finding the last element
    # so place it outside the loop
    current.next= None

Answer 2

我不确定你的 while 循环在做什么，如果你想删除最后一个元素，你会这样做：

1. 处理空列表情况（什么都不做）
2. 处理 1 元素列表案例（删除整个列表）

3. 查找倒数第二个元素，将其设置为无，最后指向该元素：

   def delete (self): #want to delete last cell
       if self.top is None:
           return None
       if self.top == self.last:
           self.top, self.last = None, None
           self.length = 0
           return
       self.length -= 1        
       current = self.top
       while current.next != self.last:
           current = current.next
       current.next = None
       self.last = current
   

---

删除一个有值的单元格。

1. 检查空列表
2. 遍历列表，跟踪前任，直到元素 被发现。
3. 如果没有找到，什么也不做。
4. 如果用户正在删除链表的头部，重置顶部
5. 找到后，将前任的下一个指向找到的下一个。

这是该算法的一个示例：

def delete_value(self, value):
    if self.top == null: # empty list
        return
    predecessor, current = None, self.top
    while current.value != value:
        predecessor = current
        current = current.next
    if current is None: # not found, nothing to do:
        return
    self.length -= 1
    if predecessor is None: # self.top == current, we are deleting the head of the linked list
        self.top = self.top.next
    else:
        predecessor.next = current.next