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。但我有删除方法的问题。我想删除列表中的最后一项

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Cell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

方法toString（）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，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，请按照以下方式进行：

处理空列表案例（什么都不做）
处理1元素列表案例（删除整个列表）

找到最后一个元素的下一个元素，将其旁边的元素设置为None，并指向该元素的最后一个元素：

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

删除具有值的单元格。

检查空列表
遍历列表，跟踪前任，直到元素找到了。
如果找不到，请不要做任何事。
如果用户正在删除链接列表的头部，请重置顶部
找到后，将前一位置放在找到的旁边。

以下是该算法的示例：

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

Python 3.从链表中删除项目

2 个答案: