我一直在尝试在Python中创建一个双链表，但我在LinkedList类中遇到了一些方法。我希望我的removeFront和removeRear方法返回已删除的值，但我无法使其工作。使用测试列表，如果我将一些值x输入节点并尝试将其删除，则不会返回x。
我认为删除节点的想法是通过将其下一个节点与其前一节点连接来将其从列表中删除，但感觉我对此的尝试存在根本缺陷。
我也在尝试实现pop方法（删除列表中的特定元素）但不知道从哪里开始。任何建议在这里向我推进正确的方向将不胜感激。谢谢。
class Node:
def __init__(self,data):
self.data = data
self.next = None
self.prev = None
class LinkedList:
def __init__(self):
self.front = None
self.rear = None
def isEmpty(self):
return self.front is None and self.rear is None
def addFront(self, data):
new_node = Node(data)
if self.front is None:
self.front = new_node
self.rear = self.front
self.front.prev = None
self.rear.next = None
else:
self.front.prev = new_node
new_node.next = self.front
self.front = new_node
self.front.prev = None
def addRear(self, data):
new_node = Node(data)
if self.front is None:
self.rear = new_node
self.front = self.rear
self.front.prev = None
self.rear.next = None
else:
self.rear.next = new_node
new_node.prev = self.rear
self.rear = new_node
self.rear.next = None
def removeFront(self):
if self.isEmpty():
return None
else:
removed = self.front
self.front.prev = self.front
self.front.prev.next = None
return removed
def removeRear(self):
if self.isEmpty():
return None
else:
removed = self.rear
self.rear.prev = self.rear
self.rear.prev.next = None
return removed
def pop(self, index):
# TODO: How to implement?
pass
def size(self):
current = self.front
count = 0
while current is not None:
count += 1
current = current.next
return count
答案 0 :(得分：1)
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的实施：
removeFront def removeFront(self):
if self.isEmpty():
return None
# Store data so that it can be returned
data = self.front.data
if self.front == self.rear:
# List contains one item, set front & rear to initial state
self.front = self.rear = None
else:
# More than one item
self.front = self.front.next # Set front to point next item
self.front.prev = None # Front doesn't have previous item
return data
完全符合相同的逻辑。
为了实施removeRear，您需要考虑三种不同的情况：
无论遇到什么情况，您仍需要找到要删除的正确节点。您可以在列表中前进的同时循环执行此操作。找到正确的节点后，如果它是第一个或最后一个节点，您可以使用pop或removeFront。如果节点介于两者之间，则需要将其从之前的＆amp;下一个节点并将剩余的节点连接在一起。
答案 1 :(得分：0)
快速代码审核：
Node.__init__()是正确的。LinkedList.__init__()是正确的。LinkedList.isEmpty()是正确的。LinkedList.size()是正确的。以下是如何正确实施LinkedList.removeFront()：
def removeFront(self):
if self.isEmpty():
return None
else:
removed = self.front.data
self.front = self.front.next
if self.front is None: # List size was 1
self.rear = None
else:
self.front.prev = None
return removed
我会将removeRear()的实施作为练习留给您。逻辑与removeFront()对称。
LinkedList.addFront()是正确的，但可以简化：
def addFront(self, data):
new_node = Node(data)
if self.front is None:
self.front = new_node
self.rear = new_node
#self.front.prev = None # Already None
#self.rear.next = None # Already None
else:
self.front.prev = new_node
new_node.next = self.front
self.front = new_node
#self.front.prev = None # Already None
您的LinkedList.addRear()是正确的，但可以采用相同的方式进行简化。
您询问了如何实施LinkedList.pop()。这将是一个棘手的问题。你必须考虑多个案例：