值为单链表的Python字典

Question

我一直在尝试创建一个值为单链表的Python字典。我的代码看起来不错，产生了正确的键。但是每个键都产生相同的值，不知道为什么。如果您有任何解决方案，请回复，代码如下。

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

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

def append(self, data):
new_node = Node(data)
if self.head is None:
self.head = new_node
else:
tail = self.head
while tail.next:
tail = tail.next
tail.next = new_node

def create_two_ll(self, lists):
ll_dict = {}
for i in range(len(lists)):
for data in lists[i]:
self.append(data)
ll_dict[f"ll{i}"] = ll
print(ll_dict)
print(ll_dict["ll0"].head.next.next.next.data)
print(ll_dict["ll1"].head.next.next.next.data)
return ll_dict

if __name__ == "__main__":
ll = LinkedList()
ll.create_two_ll([[1,3,5], [2,4,6]])

输出(错误的):

{'ll0': <__main__.LinkedList object at 0x7feead82c850>}
{'ll0': <__main__.LinkedList object at 0x7feead82c850>, 'll1': <__main__.LinkedList object at 0x7feead82c850>}
2
2

Answer 1

我用另一种方法解决了，但我不确定是否可行

我希望它能激发你更好的想法

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

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

def append(self, data):
new_node = Node(data)
if self.head is None:
self.head = new_node
else:
tail = self.head
while tail.next:
tail = tail.next
tail.next = new_node

def create_two_ll(self, lists):
ll_dict = {}
for i in range(len(lists)):
for data in lists[i]:
self.append(data)
ll_dict[f"ll{i}"] = data
print(ll_dict)
return ll_dict

if __name__ == "__main__":
ll = LinkedList()
ll.create_two_ll([[1,3,5], [2,4,6],[7,9,11]])