在python中使用链表的最简单方法是什么？在方案中，链接列表仅由'(1 2 3 4 5)定义。 Python的列表[1, 2, 3, 4, 5]和元组(1, 2, 3, 4, 5)实际上不是链表，链表有一些很好的属性，例如常量时间连接，并且能够引用它们的不同部分。让它们一成不变，它们真的很容易合作！
答案 0 :(得分：154)
对于某些需求，deque也可能有用。您可以以（1）的成本在双端队列的两端添加和删除项目。
from collections import deque
d = deque([1,2,3,4])
print d
for x in d:
print x
print d.pop(), d
答案 1 :(得分：69)
以下是一些基于Martin v. Löwis's representation的列表函数：
cons = lambda el, lst: (el, lst)
mklist = lambda *args: reduce(lambda lst, el: cons(el, lst), reversed(args), None)
car = lambda lst: lst[0] if lst else lst
cdr = lambda lst: lst[1] if lst else lst
nth = lambda n, lst: nth(n-1, cdr(lst)) if n > 0 else car(lst)
length = lambda lst, count=0: length(cdr(lst), count+1) if lst else count
begin = lambda *args: args[-1]
display = lambda lst: begin(w("%s " % car(lst)), display(cdr(lst))) if lst else w("nil\n")
其中w = sys.stdout.write
尽管Raymond Hettinger的ordered set recipe中使用了双重链表，但单链表在Python中没有实际价值。
我从未在Python中使用单链表来解决除教育之外的任何问题。
Thomas Watnedal suggested一个很好的教育资源How to Think Like a Computer Scientist, Chapter 17: Linked lists：
链接列表是：
包含货物对象和对链接列表的引用的节点。
class Node:
def __init__(self, cargo=None, next=None):
self.car = cargo
self.cdr = next
def __str__(self):
return str(self.car)
def display(lst):
if lst:
w("%s " % lst)
display(lst.cdr)
else:
w("nil\n")
答案 2 :(得分：66)
我前几天写了这篇文章
#! /usr/bin/env python
class Node(object):
def __init__(self):
self.data = None # contains the data
self.next = None # contains the reference to the next node
class LinkedList:
def __init__(self):
self.cur_node = None
def add_node(self, data):
new_node = Node() # create a new node
new_node.data = data
new_node.next = self.cur_node # link the new node to the 'previous' node.
self.cur_node = new_node # set the current node to the new one.
def list_print(self):
node = self.cur_node # cant point to ll!
while node:
print node.data
node = node.next
ll = LinkedList()
ll.add_node(1)
ll.add_node(2)
ll.add_node(3)
ll.list_print()
答案 3 :(得分：34)
接受的答案相当复杂。这是一个更标准的设计：
L = LinkedList()
L.insert(1)
L.insert(1)
L.insert(2)
L.insert(4)
print L
L.clear()
print L
这是一个简单的LinkedList类，基于简单的C ++设计和Chapter 17: Linked lists，由Thomas Watnedal推荐。
class Node:
def __init__(self, value = None, next = None):
self.value = value
self.next = next
def __str__(self):
return 'Node ['+str(self.value)+']'
class LinkedList:
def __init__(self):
self.first = None
self.last = None
def insert(self, x):
if self.first == None:
self.first = Node(x, None)
self.last = self.first
elif self.last == self.first:
self.last = Node(x, None)
self.first.next = self.last
else:
current = Node(x, None)
self.last.next = current
self.last = current
def __str__(self):
if self.first != None:
current = self.first
out = 'LinkedList [\n' +str(current.value) +'\n'
while current.next != None:
current = current.next
out += str(current.value) + '\n'
return out + ']'
return 'LinkedList []'
def clear(self):
self.__init__()
答案 4 :(得分：17)
不可变列表最好通过两元组表示，其中None表示NIL。要允许简单地制定此类列表，您可以使用此功能：
def mklist(*args):
result = None
for element in reversed(args):
result = (element, result)
return result
要使用这些列表，我宁愿提供整个LISP函数集合（即第一，第二，第n等），而不是引入方法。
答案 5 :(得分：13)
这是链表类的稍微复杂的版本，具有与python的序列类型类似的接口（即支持索引，切片，与任意序列串联等）。它应该有O（1）前置，除非需要，否则不会复制数据，并且可以与元组非常互换地使用。
它不会像lisp cons单元那样具有空间或时间效率，因为python类显然有点重量级（你可以用“__slots__ = '_head','_tail'”稍微改进一下以减少内存使用量）。然而，它将具有所需的大O性能特征。
使用示例：
>>> l = LinkedList([1,2,3,4])
>>> l
LinkedList([1, 2, 3, 4])
>>> l.head, l.tail
(1, LinkedList([2, 3, 4]))
# Prepending is O(1) and can be done with:
LinkedList.cons(0, l)
LinkedList([0, 1, 2, 3, 4])
# Or prepending arbitrary sequences (Still no copy of l performed):
[-1,0] + l
LinkedList([-1, 0, 1, 2, 3, 4])
# Normal list indexing and slice operations can be performed.
# Again, no copy is made unless needed.
>>> l[1], l[-1], l[2:]
(2, 4, LinkedList([3, 4]))
>>> assert l[2:] is l.next.next
# For cases where the slice stops before the end, or uses a
# non-contiguous range, we do need to create a copy. However
# this should be transparent to the user.
>>> LinkedList(range(100))[-10::2]
LinkedList([90, 92, 94, 96, 98])
实现：
import itertools
class LinkedList(object):
"""Immutable linked list class."""
def __new__(cls, l=[]):
if isinstance(l, LinkedList): return l # Immutable, so no copy needed.
i = iter(l)
try:
head = i.next()
except StopIteration:
return cls.EmptyList # Return empty list singleton.
tail = LinkedList(i)
obj = super(LinkedList, cls).__new__(cls)
obj._head = head
obj._tail = tail
return obj
@classmethod
def cons(cls, head, tail):
ll = cls([head])
if not isinstance(tail, cls):
tail = cls(tail)
ll._tail = tail
return ll
# head and tail are not modifiable
@property
def head(self): return self._head
@property
def tail(self): return self._tail
def __nonzero__(self): return True
def __len__(self):
return sum(1 for _ in self)
def __add__(self, other):
other = LinkedList(other)
if not self: return other # () + l = l
start=l = LinkedList(iter(self)) # Create copy, as we'll mutate
while l:
if not l._tail: # Last element?
l._tail = other
break
l = l._tail
return start
def __radd__(self, other):
return LinkedList(other) + self
def __iter__(self):
x=self
while x:
yield x.head
x=x.tail
def __getitem__(self, idx):
"""Get item at specified index"""
if isinstance(idx, slice):
# Special case: Avoid constructing a new list, or performing O(n) length
# calculation for slices like l[3:]. Since we're immutable, just return
# the appropriate node. This becomes O(start) rather than O(n).
# We can't do this for more complicated slices however (eg [l:4]
start = idx.start or 0
if (start >= 0) and (idx.stop is None) and (idx.step is None or idx.step == 1):
no_copy_needed=True
else:
length = len(self) # Need to calc length.
start, stop, step = idx.indices(length)
no_copy_needed = (stop == length) and (step == 1)
if no_copy_needed:
l = self
for i in range(start):
if not l: break # End of list.
l=l.tail
return l
else:
# We need to construct a new list.
if step < 1: # Need to instantiate list to deal with -ve step
return LinkedList(list(self)[start:stop:step])
else:
return LinkedList(itertools.islice(iter(self), start, stop, step))
else:
# Non-slice index.
if idx < 0: idx = len(self)+idx
if not self: raise IndexError("list index out of range")
if idx == 0: return self.head
return self.tail[idx-1]
def __mul__(self, n):
if n <= 0: return Nil
l=self
for i in range(n-1): l += self
return l
def __rmul__(self, n): return self * n
# Ideally we should compute the has ourselves rather than construct
# a temporary tuple as below. I haven't impemented this here
def __hash__(self): return hash(tuple(self))
def __eq__(self, other): return self._cmp(other) == 0
def __ne__(self, other): return not self == other
def __lt__(self, other): return self._cmp(other) < 0
def __gt__(self, other): return self._cmp(other) > 0
def __le__(self, other): return self._cmp(other) <= 0
def __ge__(self, other): return self._cmp(other) >= 0
def _cmp(self, other):
"""Acts as cmp(): -1 for self<other, 0 for equal, 1 for greater"""
if not isinstance(other, LinkedList):
return cmp(LinkedList,type(other)) # Arbitrary ordering.
A, B = iter(self), iter(other)
for a,b in itertools.izip(A,B):
if a<b: return -1
elif a > b: return 1
try:
A.next()
return 1 # a has more items.
except StopIteration: pass
try:
B.next()
return -1 # b has more items.
except StopIteration: pass
return 0 # Lists are equal
def __repr__(self):
return "LinkedList([%s])" % ', '.join(map(repr,self))
class EmptyList(LinkedList):
"""A singleton representing an empty list."""
def __new__(cls):
return object.__new__(cls)
def __iter__(self): return iter([])
def __nonzero__(self): return False
@property
def head(self): raise IndexError("End of list")
@property
def tail(self): raise IndexError("End of list")
# Create EmptyList singleton
LinkedList.EmptyList = EmptyList()
del EmptyList
答案 6 :(得分：6)
llist模块实现链表数据结构。它支持双向链表，即dllist和单链接的数据结构sllist。
此对象表示双向链表数据结构。
first 列表中的第一个dllistnode对象。 None如果列表为空。
last 列表中的最后dllistnode个对象。如果列表为空，则为无。
dllist对象也支持以下方法：
append(x) 将x添加到列表的右侧并返回已插入的dllistnode。
appendleft(x) 将x添加到列表的左侧并返回已插入的dllistnode。
appendright(x) 将x添加到列表的右侧并返回已插入的dllistnode。
clear() 从列表中删除所有节点。
extend(iterable) 将iterable中的元素追加到列表的右侧。
extendleft(iterable) 将iterable中的元素附加到列表的左侧。
extendright(iterable) 将iterable中的元素追加到列表的右侧。
insert(x[, before]) 如果未指定x，请将before添加到列表的右侧，或将x插入dllistnode before的左侧。返回已插入dllistnode。
nodeat(index) 在dllistnode返回节点（类型index）。
pop() 从列表的右侧删除并返回元素的值。
popleft() 从列表左侧删除并返回元素的值。
popright() 从列表右侧删除并返回元素的值
remove(node) 从列表中删除node并返回存储在其中的元素。
dllistnode个对象llist.dllistnode([value]) 返回一个新的双向链表节点，用value初始化（可选）。
dllistnode个对象提供以下属性：next 列表中的下一个节点。该属性是只读的。
prev 列表中的上一个节点。该属性是只读的。
value 存储在此节点中的值。 Compiled from this reference
班级llist.sllist([iterable])
返回使用iterable中的元素初始化的新单链表。如果未指定iterable，则新sllist为空。
为此sllist对象定义了一组类似的属性和操作。 See this reference for more information.
答案 7 :(得分：4)
class Node(object):
def __init__(self, data=None, next=None):
self.data = data
self.next = next
def setData(self, data):
self.data = data
return self.data
def setNext(self, next):
self.next = next
def getNext(self):
return self.next
def hasNext(self):
return self.next != None
class singleLinkList(object):
def __init__(self):
self.head = None
def isEmpty(self):
return self.head == None
def insertAtBeginning(self, data):
newNode = Node()
newNode.setData(data)
if self.listLength() == 0:
self.head = newNode
else:
newNode.setNext(self.head)
self.head = newNode
def insertAtEnd(self, data):
newNode = Node()
newNode.setData(data)
current = self.head
while current.getNext() != None:
current = current.getNext()
current.setNext(newNode)
def listLength(self):
current = self.head
count = 0
while current != None:
count += 1
current = current.getNext()
return count
def print_llist(self):
current = self.head
print("List Start.")
while current != None:
print(current.getData())
current = current.getNext()
print("List End.")
if __name__ == '__main__':
ll = singleLinkList()
ll.insertAtBeginning(55)
ll.insertAtEnd(56)
ll.print_llist()
print(ll.listLength())
答案 8 :(得分：2)
class Node(object):
def __init__(self):
self.data = None
self.next = None
class LinkedList:
def __init__(self):
self.head = None
def prepend_node(self, data):
new_node = Node()
new_node.data = data
new_node.next = self.head
self.head = new_node
def append_node(self, data):
new_node = Node()
new_node.data = data
current = self.head
while current.next:
current = current.next
current.next = new_node
def reverse(self):
""" In-place reversal, modifies exiting list"""
previous = None
current_node = self.head
while current_node:
temp = current_node.next
current_node.next = previous
previous = current_node
current_node = temp
self.head = previous
def search(self, data):
current_node = self.head
try:
while current_node.data != data:
current_node = current_node.next
return True
except:
return False
def display(self):
if self.head is None:
print("Linked list is empty")
else:
current_node = self.head
while current_node:
print(current_node.data)
current_node = current_node.next
def list_length(self):
list_length = 0
current_node = self.head
while current_node:
list_length += 1
current_node = current_node.next
return list_length
def main():
linked_list = LinkedList()
linked_list.prepend_node(1)
linked_list.prepend_node(2)
linked_list.prepend_node(3)
linked_list.append_node(24)
linked_list.append_node(25)
linked_list.display()
linked_list.reverse()
linked_list.display()
print(linked_list.search(1))
linked_list.reverse()
linked_list.display()
print("Lenght of singly linked list is: " + str(linked_list.list_length()))
if __name__ == "__main__":
main()
答案 9 :(得分：2)
我将此附加功能基于Nick Stinemates
def add_node_at_end(self, data):
new_node = Node()
node = self.curr_node
while node:
if node.next == None:
node.next = new_node
new_node.next = None
new_node.data = data
node = node.next
他在开始时添加新节点的方法虽然我看到很多实现通常在最后添加一个新节点但是无论如何，这很有趣。
答案 10 :(得分：2)
以下是我提出的建议。在这个帖子中，它是Riccardo C.'s的similer，除了它按顺序打印数字而不是反向打印。我还将LinkedList对象设置为Python Iterator，以便像普通的Python列表一样打印列表。
class Node:
def __init__(self, data=None):
self.data = data
self.next = None
def __str__(self):
return str(self.data)
class LinkedList:
def __init__(self):
self.head = None
self.curr = None
self.tail = None
def __iter__(self):
return self
def next(self):
if self.head and not self.curr:
self.curr = self.head
return self.curr
elif self.curr.next:
self.curr = self.curr.next
return self.curr
else:
raise StopIteration
def append(self, data):
n = Node(data)
if not self.head:
self.head = n
self.tail = n
else:
self.tail.next = n
self.tail = self.tail.next
# Add 5 nodes
ll = LinkedList()
for i in range(1, 6):
ll.append(i)
# print out the list
for n in ll:
print n
"""
Example output:
$ python linked_list.py
1
2
3
4
5
"""
答案 11 :(得分：2)
我刚做this作为一个有趣的玩具。它应该是不可变的，只要你不触及下划线前缀方法，它就会实现一堆Python魔术，比如索引和len。
答案 12 :(得分：1)
class LinkedStack:
'''LIFO Stack implementation using a singly linked list for storage.'''
_ToList = []
#---------- nested _Node class -----------------------------
class _Node:
'''Lightweight, nonpublic class for storing a singly linked node.'''
__slots__ = '_element', '_next' #streamline memory usage
def __init__(self, element, next):
self._element = element
self._next = next
#--------------- stack methods ---------------------------------
def __init__(self):
'''Create an empty stack.'''
self._head = None
self._size = 0
def __len__(self):
'''Return the number of elements in the stack.'''
return self._size
def IsEmpty(self):
'''Return True if the stack is empty'''
return self._size == 0
def Push(self,e):
'''Add element e to the top of the Stack.'''
self._head = self._Node(e, self._head) #create and link a new node
self._size +=1
self._ToList.append(e)
def Top(self):
'''Return (but do not remove) the element at the top of the stack.
Raise exception if the stack is empty
'''
if self.IsEmpty():
raise Exception('Stack is empty')
return self._head._element #top of stack is at head of list
def Pop(self):
'''Remove and return the element from the top of the stack (i.e. LIFO).
Raise exception if the stack is empty
'''
if self.IsEmpty():
raise Exception('Stack is empty')
answer = self._head._element
self._head = self._head._next #bypass the former top node
self._size -=1
self._ToList.remove(answer)
return answer
def Count(self):
'''Return how many nodes the stack has'''
return self.__len__()
def Clear(self):
'''Delete all nodes'''
for i in range(self.Count()):
self.Pop()
def ToList(self):
return self._ToList
答案 13 :(得分：1)
我在https://pypi.python.org/pypi/linked_list_mod/
上放了一个Python 2.x和3.x单链表列表使用CPython 2.7，CPython 3.4，Pypy 2.3.1，Pypy3 2.3.1和Jython 2.7b2进行了测试，并附带了一个不错的自动化测试套件。
它还包括LIFO和FIFO类。
虽然它们不可变。
答案 14 :(得分：1)
关联列表类
glPushMatrix(); {
glScalef(scale.x, scale.y, scale.z);
glTranslatef(translate.x, translate.y, translate.z);
glRotatef(rot_angle, rotation.x, rotation.y, rotation.z);
glEnableClientState(GL_VERTEX_ARRAY);
//glEnableClientState(GL_COLOR_ARRAY);
glEnableClientState(GL_NORMAL_ARRAY);
glEnableClientState(GL_TEXTURE_COORD_ARRAY);
//glColorPointer(4, GL_FLOAT, 0, colors.data());
glVertexPointer(3, GL_FLOAT, 0, verts.data());
glNormalPointer(GL_FLOAT, 0, norms.data());
glTexCoordPointer(2, GL_FLOAT, 0, texcoords.data());
glMaterialfv(GL_FRONT, GL_AMBIENT, ambient.data()); // set ambient to what is defined in scene
glMaterialfv(GL_FRONT, GL_DIFFUSE, diffuse.data()); // set diffuse to what is defined in scene
glMaterialfv(GL_FRONT, GL_SPECULAR, specular.data()); // set specular to what is defined in scene
glMaterialfv(GL_FRONT, GL_EMISSION, emission.data()); // set emission to what is defined in scene
glMaterialfv(GL_FRONT, GL_SHININESS, shininess.data()); // set shininess to what is defined in scene
glColor4f(R, G, B, A);
glBindTexture(GL_TEXTURE_2D, *texture);
glDrawArrays(primitive, 0, verts.size() / 3);
glBindTexture(GL_TEXTURE_2D, NULL);
glColor4f(1.0f, 1.0f, 1.0f, 1.0f);
glMaterialfv(GL_FRONT, GL_AMBIENT, ambient_def); // set ambient to default values
glMaterialfv(GL_FRONT, GL_DIFFUSE, diffuse_def); // set diffuse to default values
glMaterialfv(GL_FRONT, GL_SPECULAR, specular_def); // set specular to default values
glMaterialfv(GL_FRONT, GL_EMISSION, emission_def); // set emission to default values
glMaterialfv(GL_FRONT, GL_SHININESS, shininess_def); // set shininess to default value
glDisableClientState(GL_VERTEX_ARRAY);
//glDisableClientState(GL_COLOR_ARRAY);
glDisableClientState(GL_NORMAL_ARRAY);
glDisableClientState(GL_TEXTURE_COORD_ARRAY);
} glPopMatrix();
<强>用法
class LinkedStack:
# Nested Node Class
class Node:
def __init__(self, element, next):
self.__element = element
self.__next = next
def get_next(self):
return self.__next
def get_element(self):
return self.__element
def __init__(self):
self.head = None
self.size = 0
self.data = []
def __len__(self):
return self.size
def __str__(self):
return str(self.data)
def is_empty(self):
return self.size == 0
def push(self, e):
newest = self.Node(e, self.head)
self.head = newest
self.size += 1
self.data.append(newest)
def top(self):
if self.is_empty():
raise Empty('Stack is empty')
return self.head.__element
def pop(self):
if self.is_empty():
raise Empty('Stack is empty')
answer = self.head.element
self.head = self.head.next
self.size -= 1
return answer
输出
from LinkedStack import LinkedStack
x = LinkedStack()
x.push(10)
x.push(25)
x.push(55)
for i in range(x.size - 1, -1, -1):
print '|', x.data[i].get_element(), '|' ,
#next object
if x.data[i].get_next() == None:
print '--> None'
else:
print x.data[i].get_next().get_element(), '-|----> ',
答案 15 :(得分：1)
这是我的简单实现：
class Node:
def __init__(self):
self.data = None
self.next = None
def __str__(self):
return "Data %s: Next -> %s"%(self.data, self.next)
class LinkedList:
def __init__(self):
self.head = Node()
self.curNode = self.head
def insertNode(self, data):
node = Node()
node.data = data
node.next = None
if self.head.data == None:
self.head = node
self.curNode = node
else:
self.curNode.next = node
self.curNode = node
def printList(self):
print self.head
l = LinkedList()
l.insertNode(1)
l.insertNode(2)
l.insertNode(34)
输出：
Data 1: Next -> Data 2: Next -> Data 34: Next -> Data 4: Next -> None
答案 16 :(得分：1)
这是我的解决方案：
<强>实施
col1 col2_1 col2_2 col2_3 col2_4
1 1 1 1 0
2 1 0 0 1
1 1 1 0 0
<强>用法
class Node:
def __init__(self, initdata):
self.data = initdata
self.next = None
def get_data(self):
return self.data
def set_data(self, data):
self.data = data
def get_next(self):
return self.next
def set_next(self, node):
self.next = node
# ------------------------ Link List class ------------------------------- #
class LinkList:
def __init__(self):
self.head = None
def is_empty(self):
return self.head == None
def traversal(self, data=None):
node = self.head
index = 0
found = False
while node is not None and not found:
if node.get_data() == data:
found = True
else:
node = node.get_next()
index += 1
return (node, index)
def size(self):
_, count = self.traversal(None)
return count
def search(self, data):
node, _ = self.traversal(data)
return node
def add(self, data):
node = Node(data)
node.set_next(self.head)
self.head = node
def remove(self, data):
previous_node = None
current_node = self.head
found = False
while current_node is not None and not found:
if current_node.get_data() == data:
found = True
if previous_node:
previous_node.set_next(current_node.get_next())
else:
self.head = current_node
else:
previous_node = current_node
current_node = current_node.get_next()
return found
原始实施理念
答案 17 :(得分：1)
class LL(object):
def __init__(self,val):
self.val = val
self.next = None
def pushNodeEnd(self,top,val):
if top is None:
top.val=val
top.next=None
else:
tmp=top
while (tmp.next != None):
tmp=tmp.next
newNode=LL(val)
newNode.next=None
tmp.next=newNode
def pushNodeFront(self,top,val):
if top is None:
top.val=val
top.next=None
else:
newNode=LL(val)
newNode.next=top
top=newNode
def popNodeFront(self,top):
if top is None:
return
else:
sav=top
top=top.next
return sav
def popNodeEnd(self,top):
if top is None:
return
else:
tmp=top
while (tmp.next != None):
prev=tmp
tmp=tmp.next
prev.next=None
return tmp
top=LL(10)
top.pushNodeEnd(top, 20)
top.pushNodeEnd(top, 30)
pop=top.popNodeEnd(top)
print (pop.val)
答案 18 :(得分：1)
使用不可变链表时，请考虑直接使用Python的元组。
ls = (1, 2, 3, 4, 5)
def first(ls): return ls[0]
def rest(ls): return ls[1:]
它非常简单，你可以保留像len（ls），x在ls等其他功能。
答案 19 :(得分：0)
首先，我假设你想要链表。在实践中，您可以使用collections.deque，其当前的CPython实现是一个双向链接的块列表（每个块包含62个货物对象的数组）。它包含链表的功能。您还可以在pypi上搜索名为llist的C扩展名。如果你想要一个纯Python和易于理解的链表ADT实现，你可以看看我的以下最小实现。
class Node (object):
""" Node for a linked list. """
def __init__ (self, value, next=None):
self.value = value
self.next = next
class LinkedList (object):
""" Linked list ADT implementation using class.
A linked list is a wrapper of a head pointer
that references either None, or a node that contains
a reference to a linked list.
"""
def __init__ (self, iterable=()):
self.head = None
for x in iterable:
self.head = Node(x, self.head)
def __iter__ (self):
p = self.head
while p is not None:
yield p.value
p = p.next
def prepend (self, x): # 'appendleft'
self.head = Node(x, self.head)
def reverse (self):
""" In-place reversal. """
p = self.head
self.head = None
while p is not None:
p0, p = p, p.next
p0.next = self.head
self.head = p0
if __name__ == '__main__':
ll = LinkedList([6,5,4])
ll.prepend(3); ll.prepend(2)
print list(ll)
ll.reverse()
print list(ll)
答案 20 :(得分：0)
class LinkedList:
def __init__(self, value):
self.value = value
self.next = None
def insert(self, node):
if not self.next:
self.next = node
else:
self.next.insert(node)
def __str__(self):
if self.next:
return '%s -> %s' % (self.value, str(self.next))
else:
return ' %s ' % self.value
if __name__ == "__main__":
items = ['a', 'b', 'c', 'd', 'e']
ll = None
for item in items:
if ll:
next_ll = LinkedList(item)
ll.insert(next_ll)
else:
ll = LinkedList(item)
print('[ %s ]' % ll)
答案 21 :(得分：0)
双重链接列表的示例（另存为linkedlist.py）：
class node:
def __init__(self, before=None, cargo=None, next=None):
self._previous = before
self._cargo = cargo
self._next = next
def __str__(self):
return str(self._cargo) or None
class linkedList:
def __init__(self):
self._head = None
self._length = 0
def add(self, cargo):
n = node(None, cargo, self._head)
if self._head:
self._head._previous = n
self._head = n
self._length += 1
def search(self,cargo):
node = self._head
while (node and node._cargo != cargo):
node = node._next
return node
def delete(self,cargo):
node = self.search(cargo)
if node:
prev = node._previous
nx = node._next
if prev:
prev._next = node._next
else:
self._head = nx
nx._previous = None
if nx:
nx._previous = prev
else:
prev._next = None
self._length -= 1
def __str__(self):
print 'Size of linked list: ',self._length
node = self._head
while node:
print node
node = node._next
测试（另存为test.py）：
from linkedlist import node, linkedList
def test():
print 'Testing Linked List'
l = linkedList()
l.add(10)
l.add(20)
l.add(30)
l.add(40)
l.add(50)
l.add(60)
print 'Linked List after insert nodes:'
l.__str__()
print 'Search some value, 30:'
node = l.search(30)
print node
print 'Delete some value, 30:'
node = l.delete(30)
l.__str__()
print 'Delete first element, 60:'
node = l.delete(60)
l.__str__()
print 'Delete last element, 10:'
node = l.delete(10)
l.__str__()
if __name__ == "__main__":
test()
<强>输出：
Testing Linked List
Linked List after insert nodes:
Size of linked list: 6
60
50
40
30
20
10
Search some value, 30:
30
Delete some value, 30:
Size of linked list: 5
60
50
40
20
10
Delete first element, 60:
Size of linked list: 4
50
40
20
10
Delete last element, 10:
Size of linked list: 3
50
40
20
答案 22 :(得分：0)
我认为下面的实施非常优雅。
'''singly linked lists, by Yingjie Lan, December 1st, 2011'''
class linkst:
'''Singly linked list, with pythonic features.
The list has pointers to both the first and the last node.'''
__slots__ = ['data', 'next'] #memory efficient
def __init__(self, iterable=(), data=None, next=None):
'''Provide an iterable to make a singly linked list.
Set iterable to None to make a data node for internal use.'''
if iterable is not None:
self.data, self.next = self, None
self.extend(iterable)
else: #a common node
self.data, self.next = data, next
def empty(self):
'''test if the list is empty'''
return self.next is None
def append(self, data):
'''append to the end of list.'''
last = self.data
self.data = last.next = linkst(None, data)
#self.data = last.next
def insert(self, data, index=0):
'''insert data before index.
Raise IndexError if index is out of range'''
curr, cat = self, 0
while cat < index and curr:
curr, cat = curr.next, cat+1
if index<0 or not curr:
raise IndexError(index)
new = linkst(None, data, curr.next)
if curr.next is None: self.data = new
curr.next = new
def reverse(self):
'''reverse the order of list in place'''
current, prev = self.next, None
while current: #what if list is empty?
next = current.next
current.next = prev
prev, current = current, next
if self.next: self.data = self.next
self.next = prev
def delete(self, index=0):
'''remvoe the item at index from the list'''
curr, cat = self, 0
while cat < index and curr.next:
curr, cat = curr.next, cat+1
if index<0 or not curr.next:
raise IndexError(index)
curr.next = curr.next.next
if curr.next is None: #tail
self.data = curr #current == self?
def remove(self, data):
'''remove first occurrence of data.
Raises ValueError if the data is not present.'''
current = self
while current.next: #node to be examined
if data == current.next.data: break
current = current.next #move on
else: raise ValueError(data)
current.next = current.next.next
if current.next is None: #tail
self.data = current #current == self?
def __contains__(self, data):
'''membership test using keyword 'in'.'''
current = self.next
while current:
if data == current.data:
return True
current = current.next
return False
def __iter__(self):
'''iterate through list by for-statements.
return an iterator that must define the __next__ method.'''
itr = linkst()
itr.next = self.next
return itr #invariance: itr.data == itr
def __next__(self):
'''the for-statement depends on this method
to provide items one by one in the list.
return the next data, and move on.'''
#the invariance is checked so that a linked list
#will not be mistakenly iterated over
if self.data is not self or self.next is None:
raise StopIteration()
next = self.next
self.next = next.next
return next.data
def __repr__(self):
'''string representation of the list'''
return 'linkst(%r)'%list(self)
def __str__(self):
'''converting the list to a string'''
return '->'.join(str(i) for i in self)
#note: this is NOT the class lab! see file linked.py.
def extend(self, iterable):
'''takes an iterable, and append all items in the iterable
to the end of the list self.'''
last = self.data
for i in iterable:
last.next = linkst(None, i)
last = last.next
self.data = last
def index(self, data):
'''TODO: return first index of data in the list self.
Raises ValueError if the value is not present.'''
#must not convert self to a tuple or any other containers
current, idx = self.next, 0
while current:
if current.data == data: return idx
current, idx = current.next, idx+1
raise ValueError(data)
答案 23 :(得分：0)
我也确实基于某个教程编写了一个单一链接列表，该教程具有两个基本的Node和Linked List类，以及一些用于插入，删除，反向，排序等的其他方法。
这不是最好或最简单的方法，但是效果不错。
"""
????????????????
Single Linked List (SLL):
A simple object-oriented implementation of Single Linked List (SLL)
with some associated methods, such as create list, count nodes, delete nodes, and such.
????????????????
"""
class Node:
"""
Instantiates a node
"""
def __init__(self, value):
"""
Node class constructor which sets the value and link of the node
"""
self.info = value
self.link = None
class SingleLinkedList:
"""
Instantiates the SLL class
"""
def __init__(self):
"""
SLL class constructor which sets the value and link of the node
"""
self.start = None
def create_single_linked_list(self):
"""
Reads values from stdin and appends them to this list and creates a SLL with integer nodes
"""
try:
number_of_nodes = int(input("? Enter a positive integer between 1-50 for the number of nodes you wish to have in the list: "))
if number_of_nodes <= 0 or number_of_nodes > 51:
print("? The number of nodes though must be an integer between 1 to 50!")
self.create_single_linked_list()
except Exception as e:
print("? Error: ", e)
self.create_single_linked_list()
try:
for _ in range(number_of_nodes):
try:
data = int(input("? Enter an integer for the node to be inserted: "))
self.insert_node_at_end(data)
except Exception as e:
print("? Error: ", e)
except Exception as e:
print("? Error: ", e)
def count_sll_nodes(self):
"""
Counts the nodes of the linked list
"""
try:
p = self.start
n = 0
while p is not None:
n += 1
p = p.link
if n >= 1:
print(f"? The number of nodes in the linked list is {n}")
else:
print(f"? The SLL does not have a node!")
except Exception as e:
print("? Error: ", e)
def search_sll_nodes(self, x):
"""
Searches the x integer in the linked list
"""
try:
position = 1
p = self.start
while p is not None:
if p.info == x:
print(f"? YAAAY! We found {x} at position {position}")
return True
#Increment the position
position += 1
#Assign the next node to the current node
p = p.link
else:
print(f"? Sorry! We couldn't find {x} at any position. Maybe, you might want to use option 9 and try again later!")
return False
except Exception as e:
print("? Error: ", e)
def display_sll(self):
"""
Displays the list
"""
try:
if self.start is None:
print("? Single linked list is empty!")
return
display_sll = "? Single linked list nodes are: "
p = self.start
while p is not None:
display_sll += str(p.info) + "\t"
p = p.link
print(display_sll)
except Exception as e:
print("? Error: ", e)
def insert_node_in_beginning(self, data):
"""
Inserts an integer in the beginning of the linked list
"""
try:
temp = Node(data)
temp.link = self.start
self.start = temp
except Exception as e:
print("? Error: ", e)
def insert_node_at_end(self, data):
"""
Inserts an integer at the end of the linked list
"""
try:
temp = Node(data)
if self.start is None:
self.start = temp
return
p = self.start
while p.link is not None:
p = p.link
p.link = temp
except Exception as e:
print("? Error: ", e)
def insert_node_after_another(self, data, x):
"""
Inserts an integer after the x node
"""
try:
p = self.start
while p is not None:
if p.info == x:
break
p = p.link
if p is None:
print(f"? Sorry! {x} is not in the list.")
else:
temp = Node(data)
temp.link = p.link
p.link = temp
except Exception as e:
print("? Error: ", e)
def insert_node_before_another(self, data, x):
"""
Inserts an integer before the x node
"""
try:
# If list is empty
if self.start is None:
print("? Sorry! The list is empty.")
return
# If x is the first node, and new node should be inserted before the first node
if x == self.start.info:
temp = Node(data)
temp.link = p.link
p.link = temp
# Finding the reference to the prior node containing x
p = self.start
while p.link is not None:
if p.link.info == x:
break
p = p.link
if p.link is not None:
print(f"? Sorry! {x} is not in the list.")
else:
temp = Node(data)
temp.link = p.link
p.link = temp
except Exception as e:
print("? Error: ", e)
def insert_node_at_position(self, data, k):
"""
Inserts an integer in k position of the linked list
"""
try:
# if we wish to insert at the first node
if k == 1:
temp = Node(data)
temp.link = self.start
self.start = temp
return
p = self.start
i = 1
while i < k-1 and p is not None:
p = p.link
i += 1
if p is None:
print(f"? The max position is {i}")
else:
temp = Node(data)
temp.link = self.start
self.start = temp
except Exception as e:
print("? Error: ", e)
def delete_a_node(self, x):
"""
Deletes a node of a linked list
"""
try:
# If list is empty
if self.start is None:
print("? Sorry! The list is empty.")
return
# If there is only one node
if self.start.info == x:
self.start = self.start.link
# If more than one node exists
p = self.start
while p.link is not None:
if p.link.info == x:
break
p = p.link
if p.link is None:
print(f"? Sorry! {x} is not in the list.")
else:
p.link = p.link.link
except Exception as e:
print("? Error: ", e)
def delete_sll_first_node(self):
"""
Deletes the first node of a linked list
"""
try:
if self.start is None:
return
self.start = self.start.link
except Exception as e:
print("? Error: ", e)
def delete_sll_last_node(self):
"""
Deletes the last node of a linked list
"""
try:
# If the list is empty
if self.start is None:
return
# If there is only one node
if self.start.link is None:
self.start = None
return
# If there is more than one node
p = self.start
# Increment until we find the node prior to the last node
while p.link.link is not None:
p = p.link
p.link = None
except Exception as e:
print("? Error: ", e)
def reverse_sll(self):
"""
Reverses the linked list
"""
try:
prev = None
p = self.start
while p is not None:
next = p.link
p.link = prev
prev = p
p = next
self.start = prev
except Exception as e:
print("? Error: ", e)
def bubble_sort_sll_nodes_data(self):
"""
Bubble sorts the linked list on integer values
"""
try:
# If the list is empty or there is only one node
if self.start is None or self.start.link is None:
print("? The list has no or only one node and sorting is not required.")
end = None
while end != self.start.link:
p = self.start
while p.link != end:
q = p.link
if p.info > q.info:
p.info, q.info = q.info, p.info
p = p.link
end = p
except Exception as e:
print("? Error: ", e)
def bubble_sort_sll(self):
"""
Bubble sorts the linked list
"""
try:
# If the list is empty or there is only one node
if self.start is None or self.start.link is None:
print("? The list has no or only one node and sorting is not required.")
end = None
while end != self.start.link:
r = p = self.start
while p.link != end:
q = p.link
if p.info > q.info:
p.link = q.link
q.link = p
if p != self.start:
r.link = q.link
else:
self.start = q
p, q = q, p
r = p
p = p.link
end = p
except Exception as e:
print("? Error: ", e)
def sll_has_cycle(self):
"""
Tests the list for cycles using Tortoise and Hare Algorithm (Floyd's cycle detection algorithm)
"""
try:
if self.find_sll_cycle() is None:
return False
else:
return True
except Exception as e:
print("? Error: ", e)
def find_sll_cycle(self):
"""
Finds cycles in the list, if any
"""
try:
# If there is one node or none, there is no cycle
if self.start is None or self.start.link is None:
return None
# Otherwise,
slowR = self.start
fastR = self.start
while slowR is not None and fastR is not None:
slowR = slowR.link
fastR = fastR.link.link
if slowR == fastR:
return slowR
return None
except Exception as e:
print("? Error: ", e)
def remove_cycle_from_sll(self):
"""
Removes the cycles
"""
try:
c = self.find_sll_cycle()
# If there is no cycle
if c is None:
return
print(f"? There is a cycle at node: ", c.info)
p = c
q = c
len_cycles = 0
while True:
len_cycles += 1
q = q.link
if p == q:
break
print(f"? The cycle length is {len_cycles}")
len_rem_list = 0
p = self.start
while p != q:
len_rem_list += 1
p = p.link
q = q.link
print(f"? The number of nodes not included in the cycle is {len_rem_list}")
length_list = len_rem_list + len_cycles
print(f"? The SLL length is {length_list}")
# This for loop goes to the end of the SLL, and set the last node to None and the cycle is removed.
p = self.start
for _ in range(length_list-1):
p = p.link
p.link = None
except Exception as e:
print("? Error: ", e)
def insert_cycle_in_sll(self, x):
"""
Inserts a cycle at a node that contains x
"""
try:
if self.start is None:
return False
p = self.start
px = None
prev = None
while p is not None:
if p.info == x:
px = p
prev = p
p = p.link
if px is not None:
prev.link = px
else:
print(f"? Sorry! {x} is not in the list.")
except Exception as e:
print("? Error: ", e)
def merge_using_new_list(self, list2):
"""
Merges two already sorted SLLs by creating new lists
"""
merge_list = SingleLinkedList()
merge_list.start = self._merge_using_new_list(self.start, list2.start)
return merge_list
def _merge_using_new_list(self, p1, p2):
"""
Private method of merge_using_new_list
"""
if p1.info <= p2.info:
Start_merge = Node(p1.info)
p1 = p1.link
else:
Start_merge = Node(p2.info)
p2 = p2.link
pM = Start_merge
while p1 is not None and p2 is not None:
if p1.info <= p2.info:
pM.link = Node(p1.info)
p1 = p1.link
else:
pM.link = Node(p2.info)
p2 = p2.link
pM = pM.link
#If the second list is finished, yet the first list has some nodes
while p1 is not None:
pM.link = Node(p1.info)
p1 = p1.link
pM = pM.link
#If the second list is finished, yet the first list has some nodes
while p2 is not None:
pM.link = Node(p2.info)
p2 = p2.link
pM = pM.link
return Start_merge
def merge_inplace(self, list2):
"""
Merges two already sorted SLLs in place in O(1) of space
"""
merge_list = SingleLinkedList()
merge_list.start = self._merge_inplace(self.start, list2.start)
return merge_list
def _merge_inplace(self, p1, p2):
"""
Merges two already sorted SLLs in place in O(1) of space
"""
if p1.info <= p2.info:
Start_merge = p1
p1 = p1.link
else:
Start_merge = p2
p2 = p2.link
pM = Start_merge
while p1 is not None and p2 is not None:
if p1.info <= p2.info:
pM.link = p1
pM = pM.link
p1 = p1.link
else:
pM.link = p2
pM = pM.link
p2 = p2.link
if p1 is None:
pM.link = p2
else:
pM.link = p1
return Start_merge
def merge_sort_sll(self):
"""
Sorts the linked list using merge sort algorithm
"""
self.start = self._merge_sort_recursive(self.start)
def _merge_sort_recursive(self, list_start):
"""
Recursively calls the merge sort algorithm for two divided lists
"""
# If the list is empty or has only one node
if list_start is None or list_start.link is None:
return list_start
# If the list has two nodes or more
start_one = list_start
start_two = self._divide_list(self_start)
start_one = self._merge_sort_recursive(start_one)
start_two = self._merge_sort_recursive(start_two)
start_merge = self._merge_inplace(start_one, start_two)
return start_merge
def _divide_list(self, p):
"""
Divides the linked list into two almost equally sized lists
"""
# Refers to the third nodes of the list
q = p.link.link
while q is not None and p is not None:
# Increments p one node at the time
p = p.link
# Increments q two nodes at the time
q = q.link.link
start_two = p.link
p.link = None
return start_two
def concat_second_list_to_sll(self, list2):
"""
Concatenates another SLL to an existing SLL
"""
# If the second SLL has no node
if list2.start is None:
return
# If the original SLL has no node
if self.start is None:
self.start = list2.start
return
# Otherwise traverse the original SLL
p = self.start
while p.link is not None:
p = p.link
# Link the last node to the first node of the second SLL
p.link = list2.start
def test_merge_using_new_list_and_inplace(self):
"""
"""
LIST_ONE = SingleLinkedList()
LIST_TWO = SingleLinkedList()
LIST_ONE.create_single_linked_list()
LIST_TWO.create_single_linked_list()
print("1️⃣ The unsorted first list is: ")
LIST_ONE.display_sll()
print("2️⃣ The unsorted second list is: ")
LIST_TWO.display_sll()
LIST_ONE.bubble_sort_sll_nodes_data()
LIST_TWO.bubble_sort_sll_nodes_data()
print("1️⃣ The sorted first list is: ")
LIST_ONE.display_sll()
print("2️⃣ The sorted second list is: ")
LIST_TWO.display_sll()
LIST_THREE = LIST_ONE.merge_using_new_list(LIST_TWO)
print("The merged list by creating a new list is: ")
LIST_THREE.display_sll()
LIST_FOUR = LIST_ONE.merge_inplace(LIST_TWO)
print("The in-place merged list is: ")
LIST_FOUR.display_sll()
def test_all_methods(self):
"""
Tests all methods of the SLL class
"""
OPTIONS_HELP = """
?????????????????????????????????????????
Select a method from 1-19:
?????????????????????????????????????????
ℹ️ (1) ? Create a single liked list (SLL).
ℹ️ (2) ? Display the SLL.
ℹ️ (3) ? Count the nodes of SLL.
ℹ️ (4) ? Search the SLL.
ℹ️ (5) ? Insert a node at the beginning of the SLL.
ℹ️ (6) ? Insert a node at the end of the SLL.
ℹ️ (7) ? Insert a node after a specified node of the SLL.
ℹ️ (8) ? Insert a node before a specified node of the SLL.
ℹ️ (9) ? Delete the first node of SLL.
ℹ️ (10) ? Delete the last node of the SLL.
ℹ️ (11) ? Delete a node you wish to remove.
ℹ️ (12) ? Reverse the SLL.
ℹ️ (13) ? Bubble sort the SLL by only exchanging the integer values.
ℹ️ (14) ? Bubble sort the SLL by exchanging links.
ℹ️ (15) ? Merge sort the SLL.
ℹ️ (16) ? Insert a cycle in the SLL.
ℹ️ (17) ? Detect if the SLL has a cycle.
ℹ️ (18) ? Remove cycle in the SLL.
ℹ️ (19) ? Test merging two bubble-sorted SLLs.
ℹ️ (20) ? Concatenate a second list to the SLL.
ℹ️ (21) ? Exit.
?????????????????????????????????????????
"""
self.create_single_linked_list()
while True:
print(OPTIONS_HELP)
UI_OPTION = int(input("? Enter an integer for the method you wish to run using the above help: "))
if UI_OPTION == 1:
data = int(input("? Enter an integer to be inserted at the end of the list: "))
x = int(input("? Enter an integer to be inserted after that: "))
self.insert_node_after_another(data, x)
elif UI_OPTION == 2:
self.display_sll()
elif UI_OPTION == 3:
self.count_sll_nodes()
elif UI_OPTION == 4:
data = int(input("? Enter an integer to be searched: "))
self.search_sll_nodes(data)
elif UI_OPTION == 5:
data = int(input("? Enter an integer to be inserted at the beginning: "))
self.insert_node_in_beginning(data)
elif UI_OPTION == 6:
data = int(input("? Enter an integer to be inserted at the end: "))
self.insert_node_at_end(data)
elif UI_OPTION == 7:
data = int(input("? Enter an integer to be inserted: "))
x = int(input("? Enter an integer to be inserted before that: "))
self.insert_node_before_another(data, x)
elif UI_OPTION == 8:
data = int(input("? Enter an integer for the node to be inserted: "))
k = int(input("? Enter an integer for the position at which you wish to insert the node: "))
self.insert_node_before_another(data, k)
elif UI_OPTION == 9:
self.delete_sll_first_node()
elif UI_OPTION == 10:
self.delete_sll_last_node()
elif UI_OPTION == 11:
data = int(input("? Enter an integer for the node you wish to remove: "))
self.delete_a_node(data)
elif UI_OPTION == 12:
self.reverse_sll()
elif UI_OPTION == 13:
self.bubble_sort_sll_nodes_data()
elif UI_OPTION == 14:
self.bubble_sort_sll()
elif UI_OPTION == 15:
self.merge_sort_sll()
elif UI_OPTION == 16:
data = int(input("? Enter an integer at which a cycle has to be formed: "))
self.insert_cycle_in_sll(data)
elif UI_OPTION == 17:
if self.sll_has_cycle():
print("? The linked list has a cycle. ")
else:
print("? YAAAY! The linked list does not have a cycle. ")
elif UI_OPTION == 18:
self.remove_cycle_from_sll()
elif UI_OPTION == 19:
self.test_merge_using_new_list_and_inplace()
elif UI_OPTION == 20:
list2 = self.create_single_linked_list()
self.concat_second_list_to_sll(list2)
elif UI_OPTION == 21:
break
else:
print("? Option must be an integer, between 1 to 21.")
print()
if __name__ == '__main__':
# Instantiates a new SLL object
SLL_OBJECT = SingleLinkedList()
SLL_OBJECT.test_all_methods()
答案 24 :(得分：-1)
enter code here
enter code here
class node:
def __init__(self):
self.data = None
self.next = None
class linked_list:
def __init__(self):
self.cur_node = None
self.head = None
def add_node(self,data):
new_node = node()
if self.head == None:
self.head = new_node
self.cur_node = new_node
new_node.data = data
new_node.next = None
self.cur_node.next = new_node
self.cur_node = new_node
def list_print(self):
node = self.head
while node:
print (node.data)
node = node.next
def delete(self):
node = self.head
next_node = node.next
del(node)
self.head = next_node
a = linked_list()
a.add_node(1)
a.add_node(2)
a.add_node(3)
a.add_node(4)
a.delete()
a.list_print()
答案 25 :(得分：-1)
我的2美分
class Node:
def __init__(self, value=None, next=None):
self.value = value
self.next = next
def __str__(self):
return str(self.value)
class LinkedList:
def __init__(self):
self.first = None
self.last = None
def add(self, x):
current = Node(x, None)
try:
self.last.next = current
except AttributeError:
self.first = current
self.last = current
else:
self.last = current
def print_list(self):
node = self.first
while node:
print node.value
node = node.next
ll = LinkedList()
ll.add("1st")
ll.add("2nd")
ll.add("3rd")
ll.add("4th")
ll.add("5th")
ll.print_list()
# Result:
# 1st
# 2nd
# 3rd
# 4th
# 5th
答案 26 :(得分：-1)
我的双重链接列表对于noobies来说可能是可以理解的。 如果您熟悉C中的DS，那么这是非常易读的。
# LinkedList..
class node:
def __init__(self): ##Cluster of Nodes' properties
self.data=None
self.next=None
self.prev=None
class linkedList():
def __init__(self):
self.t = node() // for future use
self.cur_node = node() // current node
self.start=node()
def add(self,data): // appending the LL
self.new_node = node()
self.new_node.data=data
if self.cur_node.data is None:
self.start=self.new_node //For the 1st node only
self.cur_node.next=self.new_node
self.new_node.prev=self.cur_node
self.cur_node=self.new_node
def backward_display(self): //Displays LL backwards
self.t=self.cur_node
while self.t.data is not None:
print(self.t.data)
self.t=self.t.prev
def forward_display(self): //Displays LL Forward
self.t=self.start
while self.t.data is not None:
print(self.t.data)
self.t=self.t.next
if self.t.next is None:
print(self.t.data)
break
def main(self): //This is kind of the main
function in C
ch=0
while ch is not 4: //Switch-case in C
ch=int(input("Enter your choice:"))
if ch is 1:
data=int(input("Enter data to be added:"))
ll.add(data)
ll.main()
elif ch is 2:
ll.forward_display()
ll.main()
elif ch is 3:
ll.backward_display()
ll.main()
else:
print("Program ends!!")
return
ll=linkedList()
ll.main()
虽然可以在此代码中添加更多简化，但我认为原始实现会让我更容易理解。
答案 27 :(得分：-1)
Python 中链表的当前实现需要创建一个单独的类，称为 Node，以便它们可以使用主链表类连接。在提供的实现中，链表是在没有为节点定义单独的类的情况下创建的。使用建议的实现，链表更容易理解，并且可以使用打印功能简单地可视化。
class Linkedlist:
def __init__(self):
self.outer = None
def add_outermost(self, dt):
self.outer = [dt, self.outer]
def add_innermost(self, dt):
p = self.outer
if not p:
self.outer = [dt, None]
return
while p[1]:
p = p[1]
p[1] = [dt, None]
def visualize(self):
p = self.outer
l = 'Linkedlist: '
while p:
l += (str(p[0])+'->')
p = p[1]
print(l + 'None')
ll = Linkedlist()
ll.add_innermost(8)
ll.add_outermost(3)
ll.add_outermost(5)
ll.add_outermost(2)
ll.add_innermost(7)
print(ll.outer)
ll.visualize()
答案 28 :(得分：-4)
如果您只想创建一个简单的喜欢列表，请参阅此代码
1 = [1，[2，[3，[4，[5，[6，[7，[8，[9，[10]]]]]]]]]]
用于可视化此鳕鱼的执行访问http://www.pythontutor.com/visualize.html#mode=edit