Question

为 LinkedList 添加类似 Python 自带列表实现 list 中的 append, pop, insert 方法

1.为 LinkedList 添加类似 Python 自带列表实现 list 中的 append, pop, insert 方法. 分别给出你实现的这三个方法的时间复杂度.
class Node:
def __init__(self,data):
self.data = data
self.next = None

def getData(self):
return self.data

def getNext(self):
return self.next

def setData(self,newdata):
self.data = newdata

def setNext(self,newnext):
self.next = newnext

class LinkedList:

def __init__(self):
self.head = None

def isEmpty(self):
return self.head == None

def add(self,item):
temp = Node(item)
temp.setNext(self.head)
self.head = temp

def size(self):
current = self.head
count = 0
while current != None:
count += 1
current = current.getNext()

return count

def search(self,item):
current = self.head
found = False
while current != None and not found:
if current.getData() == item:
found = True
else:
current = current.getNext()

return found

def remove(self,item):
current = self.head
previous = None
found = False
while current != None and not found:
if current.getData() == item:
found = True
else:
previous = current
current = current.getNext()
if found:
if previous == None:
self.head = current.getNext()
else:
previous.setNext(current.getNext())
2.在之前的一题中, 你很可能实现的是一个复杂度为 $O(n)$ 的 append 方法. 如何修改程序, 使得可以使其复杂度为 $O(1)$? (提示: 为 LinkedList 添加一个指向链表尾部的属性.)

Answer 1

写了个insert 

def insert(self,index,item):
        previous=None
        current=self.head
        count=0
        temp=Node(item)
        if index>self.size():
            print "out index"
        elif index==0:
            temp.setNext(current)
            self.head=temp
        else:
            while index:
                index-=1
                previous=current
                current=current.getNext()
            previous.setNext(temp)
            temp.setNext(current)

追问

请问前两个的要怎么写

追答

自己想啊，都差不多的过程。就算我告诉你你以后还是不会写。

追问

def append(self item)
if self._head is None:
self._head = LNode(elem)
return
p = self._head
while p. next is not None:
p. =p.next
p.next. = LNone(elem)是不是按这个模板写。我是自学的，还是不太懂