我正在使用链接列表来存储由3个属性id，bookName和authorName组成的集合作为添加图书的参数。对象在“Book”类中定义。
代码将像库中的工具架一样工作，将3种类型的信息id，BookName和AuthorName存储到“Book”类的Linked List节点中。< / p>
我现在正在研究LinkedList类中3个不完整的方法。
他们是：
AddBookToPosition（newBook，n） - 将Book添加到所需位置'n'（例如2） DeleteBookAtPosition（n） - 将书籍删除到所需位置'n'（例如3） SortByAuthorName（） - 将按升序对存储在链接列表中的作者姓名进行排序。
问题：
编译时没有错误，但是 AddBookAtPosition 不起作用，并且不显示在输出中添加的Book。此外， DeleteBookAtPosition 也无法正常工作。
我需要帮助（对于AddBookAtPosition）和删除（DeleteBookAtPosition），它们将分别添加或删除链接列表位置n处的节点，例如程序从位置0,1,2开始按顺序显示书籍，等等
欣赏建议。
class Node :
def __init__(self, data=None, next=None):
self.data = data
self.next = None
class Book :
def __init__(self,n,id,bookName,authorName):
self.n = n
self.id = id
self.bookName = bookName
self.authorName = authorName
def print(self):
print("Position: {}".format(self.n))
print("Book ID: {}".format(self.id))
print("Book Name: {}".format(self.bookName))
print("Author Name: {}".format(self.authorName))
class LinkedList :
def __init__(self):
self.head = None
def __iter__(self):
node = self.head
while node is not None:
yield node
node = node.next
def __len__(self):
return len(list(self))
def __getitem__(self, i):
node = self.head
if node is None:
raise IndexError('LinkedList index out of range')
for n in range(i-1):
node = node.next
if node is None:
raise IndexError('LinkedList index out of range')
return node
#this will add the book to the top of the list
def AddBookToFront (self, newBook):
new_node = Node(newBook)
new_node.next = self.head
self.head = new_node
def AddBookAtPosition (self, newBook, n):
counter = 1
if n == 0:
newBook.setNext(self.head)
self.head = newBook
else:
node = self.head
while node.next is not None :
if counter == n :
newBook.setNext(node.next)
node.setNext(newBook)
return
node = node.next
counter = counter + 1
def DeleteBookAtPosition(self, n):
# If linked list is empty
if self.head == None:
return
# Store head node
temp = self.head
# If head needs to be removed
if n == 0:
self.head = temp.next
temp = None
return
# Find previous node of the node to be deleted
for i in range(n -1 ):
temp = temp.next
if temp is None:
break
# If position is more than number of nodes
if temp is None:
return
if temp.next is None:
return
# Node temp.next is the node to be deleted
# store pointer to the next of node to be deleted
next = temp.next.next
# Unlink the node from linked list
temp.next = None
temp.next = next
def __delitem__(self, i):
if self.head is None:
raise IndexError('LinkedList index out of range')
if i == 0:
self.head = self.head.next
else:
node = self.head
for n in range(i-1):
if node.next is None:
raise IndexError('LinkedList index out of range')
node = node.next
if node.next is None:
raise IndexError('LinkedList index out of range')
node.next = node.next.next
BookList = LinkedList()
BookList.AddBookToFront(Book(1, "J.R.R. Tolkien", "Lord of the Rings"))
BookList.AddBookToFront(Book(2, "Lewis Carroll", "Alice in Wonderland"))
BookList.AddBookAtPosition(Book(3, "Star Wars: Aftermath", "Chuck Wendig"), 3)
for book in BookList:
print(book.data.n)
print(book.data.id)
print(book.data.bookName)
print(book.data.authorName)
当前输出
2
Lewis Carroll
Alice in Wonderland
1
J.R.R. Tolkien
Lord of the Rings
所需输出 - 按位置按升序显示（1,2,3）
Position : 1
Book ID : 3
Book Name: Star Wars: Aftermath
Author Name: Chuck Wendig
Position : 2
Book ID : 1
Book Name: Lord of the Rings
Author Name: J.R.R. Tolkien
答案 0 :(得分：0)
为什么书既有self.n又有self.id？这本书本身不应该知道它的立场。链接列表节点都不应该知道它的位置。链接列表的主要功能，只需在插入新元素时修改一个现有元素。如果您在Book / Node中存储位置，则会消除所有好处，因为一旦插入链接列表，必须更新所有后续元素。
首先，以这种方式修改书籍：
class Book :
def __init__(self, id, bookName, authorName):
self.id = id
self.bookName = bookName
self.authorName = authorName
您可以实现自己的LinkedList.iteritems（）方法，该方法将在您遍历节点时返回节点位置：
def iteritems(self):
pos = 0
for book in self:
yield pos, book
pos += 1
请注意，如果在迭代时修改链表，它将显示错误值。用法示例：
BookList = LinkedList()
# you've mixed up book name and author name
BookList.AddBookToFront(Book(1, "J.R.R. Tolkien", "Lord of the Rings"))
BookList.AddBookToFront(Book(2, "Lewis Carroll", "Alice in Wonderland"))
BookList.AddBookToFront(Book(3, "Donald Knuth", "C programming language"))
BookList.AddBookAtPosition(Book(4, "Star Wars: Aftermath", "Chuck Wendig"), 1)
for position, book in BookList.iteritems():
print("Position: %s" % position)
print("Book ID: %s" % book.data.id)
print("Book Name: %s" % book.data.bookName)
print("Author Name: %s" % book.data.authorName)
输出：
Position: 0
Book ID: 3
Book Name: Donald Knuth
Author name: C programming language
Position: 1
Book ID: 4
Book Name: Star Wars: Aftermath
Author name: Chuck Wendig
Position: 2
Book ID: 2
Book Name: Lewis Carroll
Author name: Alice in Wonderland
Position: 3
Book ID: 1
Book Name: J.R.R. Tolkien
Author name: Lord of the Rings
A还修复了“AddBookAtPosition”，因为它没有用。它使用“新书”而不是“新节点”：
def AddBookAtPosition(self, newBook, n):
new_node = Node(newBook)
counter = 1
if n == 0:
new_node.next = self.head
self.head = new_node
else:
node = self.head
while node.next is not None :
if counter == n :
new_node.next = node.next
node.next = new_node
return
node = node.next
counter = counter + 1
答案 1 :(得分：0)
我认为python列表会为你完成工作。 （除非你说明为什么需要LL）
class Book :
def __init__(self,id,bookName,authorName):
self.id = id
self.bookName = bookName
self.authorName = authorName
def print(self):
print("Book ID: {}".format(self.id))
print("Book Name: {}".format(self.bookName))
print("Author Name: {}".format(self.authorName))
BookList = list()
#to add book object using index
BookList.insert(0,Book(1, "J.R.R. Tolkien", "Lord of the Rings"))
BookList.insert(1,Book(2, "Lewis Carroll", "Alice in Wonderland"))
BookList.insert(2,Book(3, "Star Wars: Aftermath", "Chuck Wendig"))
# to remove object from list using index
del BookList[1]
#print whole list
for book in BookList:
print(book.id)
print(book.bookName)
print(book.authorName)