问题
I am implementing a LinkedList in python(3.7.4) and the code of the module is below :-
LinkedList.py
class Node:
def __init__(self,value):
self.value = value
self.ref = None
class LinkedList(Node):
def __init__(self):
self.__head = None
self.__cur = None
self.__count = 0
def add(self,value):
if self.__head is None:
self.__cur = Node(value)
self.__head = self.__cur
else:
self.__cur.ref = Node(value)
self.__cur = self.__cur.ref
self.__count += 1
def getList(self):
temp = self.__head
while temp!=None:
yield temp.value
temp = temp.ref
def delete(self,value):
temp = self.__head
while temp!=None:
if temp.value == value and temp == self.__head:
self.__head = temp.ref
del temp
self.__count -= 1
break
elif temp.ref != None and temp.ref.value == value:
temp_ref = temp.ref.ref
del temp.ref
self.__count -= 1
temp.ref = temp_ref
break
temp = temp.ref
def __getitem__(self,index):
i = 0
temp = self.__head
if type(index) is int:
while temp!=None:
if i == index:
return temp.value
temp = temp.ref
i += 1
elif type(index) is slice:
if index.start is None:
start = 0
else: start = index.start
if index.stop is None:
stop = self.__count
else: stop = index.stop
if index.step is None:
step = 1
else: step = index.step
returningList = list()
while temp!=None:
if start <= i < stop:
returningList.append(temp.value)
if i==0:
i = start
for _ in range(start):
if temp != None:
temp = temp.ref
else:
i+=step
for _ in range(step):
if temp != None:
temp = temp.ref
return returningList
def __len__(self):
return self.__count
All the above functions are working well, there is no any error in this module.
but my problem is __getitem__() method. I am unable to make the exact logic for that and it is going too larger.
also it is not working for negative indices like obj[-1] returning me nothing ( len(obj) is not 0 here).
can anyone give or suggest me proper logic for __getitem__() method for code optimization and complexity reduction.
回答1:
You can do this, for example:
def __getitem__(self, index):
if isinstance(index, int):
if index < 0:
index = len(self) + index
# check if `index` is valid
# search for the element as you're currently doing.
elif isinstance(index, slice):
return [self[i] for i in range(len(self))[index]]
else:
raise ValueError(f'Linked list cannot be indexed with values of type {type(index)}')
UPDATE: the code above is very concise, but it's also tremendously slow. If I'm not mistaken, it's a bit better than O(n**2), while the code below is at least 71.58 times faster (doing linkedListWith500Elements[::-1]), and it should be about O(n)!
This should be way faster because it doesn't iterate through the list each time to retrieve the next element of the slice:
class LinkedList:
...
def __iter__(self):
temp = self.__head
while temp is not None:
yield temp.value
temp = temp.ref
def __getitem__(self, index):
if isinstance(index, int):
if index < 0:
index = len(self) + index
for i, value in enumerate(self):
if i == index:
return value
raise IndexError(f'{type(self).__name__} index {index} out of range(0, {len(self)})')
elif isinstance(index, slice):
rangeOfIndices = range(len(self))[index]
isRangeIncreasing = rangeOfIndices.start <= rangeOfIndices.stop + 1 and rangeOfIndices.step > 0
rangeOfIndices = iter(rangeOfIndices) if isRangeIncreasing else reversed(rangeOfIndices)
retval = [] # you can preallocate this...
updateRetval = retval.append if isRangeIncreasing else (lambda value: retval.insert(0, value)) # ...and change this accordingly, although I haven't tested whether it'll be faster
try:
searchingForIndex = next(rangeOfIndices)
except StopIteration:
return retval
temp = self.__head
for i, element in enumerate(self):
if temp is None:
break
if i == searchingForIndex:
updateRetval(temp.value)
try:
searchingForIndex = next(rangeOfIndices)
except StopIteration:
return retval
temp = temp.ref
return retval
raise ValueError(f'{type(self).__name__} can only be indexed with integers or slices (not {type(index)})')
Preallocating the list should be around 22% faster:
...
rangeOfIndices = range(len(self))[index]
isRangeIncreasing = rangeOfIndices.start <= rangeOfIndices.stop + 1 and rangeOfIndices.step > 0
# preallocate the list...
retval = [None] * len(rangeOfIndices)
if isRangeIncreasing:
retvalIndex = 0
rangeOfIndices = iter(rangeOfIndices)
# ...and use a different update function
def updateRetval(value):
nonlocal retvalIndex
retval[retvalIndex] = value
retvalIndex += 1
else:
retvalIndex = len(retval) - 1
rangeOfIndices = reversed(rangeOfIndices)
def updateRetval(value):
nonlocal retvalIndex
retval[retvalIndex] = value
retvalIndex -= 1
try:
...
回答2:
To solve this with the least amount of code, you can first convert your linked list into a python list and then slice that python list.
But first you have to rename your method getList(self) to __iter__(self). Thats the canonical name anyways.
Now __getitem__ becomes one line:
def __getitem__(self, index):
return list(self)[index]
This is not very space efficient, since it duplicates your list.
Edit:
Here is a more efficient solution. I assume again that getList(self) is renamed to __iter__.
def __getitem__(self, index):
# creates all requested indices or just one int
indices = range(len(self))[index] # constant time and space
if isinstance(indices, int):
return next(value for i, value in enumerate(self) if i == indices) # takes time proportional to the index but constant space
else:
# the double reversing is needed if the step is negative
reverse = -1 if index.step < 0 else 1 # constant time
return [value for i,value in enumerate(self) if i in indices[::reverse]][::reverse] # O(n) time and O(len(indices)) space
This is efficient, because testing if an int is in a range and slicing a range takes constant time.
回答3:
I have implemented a recursive function to solve this problem with an extra functional parameter which will show the position of current cursor to save time in over and over start tracing from first node each and every time when recursion called.
I am not saying that other answers are wrong but I used to implement principle of KISS in my code and other answers were hard to understand.
My edited code for __getitem__() is below:
class LinkedList(Node):
...
def __getitem__(self,index,fromNode=None):
global i,temp
if fromNode is None:
i,temp = 0,self.__head
if isinstance(index, int):
if index<0: index+=len(self)
while temp!=None:
if i == index:
return temp.value
temp = temp.ref
i += 1
elif isinstance(index, slice):
step = 1 if index.step is None else index.step
start = 0 if index.start is None else index.start
stop=len(self) if index.stop is None else index.stop
if start < 0: start += len(self)
elif step> 0 >stop: stop += len(self)
reverseFlag = True if step<0 else False
if reverseFlag:
trace = iter(reversed(range(len(self))[start:stop:step]))
else:
trace = iter(range(len(self))[start:stop:step])
returningList = [self.__getitem__(indice,temp) for indice in trace]
return list(reversed(returningList)) if reverseFlag else returningList
...
Can I further reduce this code or can reduce more complexity from it? If yes then please suggest me via your answer or comments.
来源：https://stackoverflow.com/questions/57400399/implementation-of-linkedlist-in-python-getitem-method