我是否在链接列表中使用正确的 append function，使用 python？

Question

I am learning linked lists by doing assignments, and the code for making a Linked list is working fine. But, manipulating it is not working properly. So, I have a Linked list with a head and I created a function that takes two arguments (head, how many the last numbers you want to append in the first). The purpose of the function is, suppose you have a linked list. 1 -> 4->5->6->2->3->9 I want the last three numbers to be on the first and the remaining first 4 numbers to come at the last of these three numbers like this. 2->3->9->1 -> 4->5->6 So, what should I need is to have broken the link between 2 and its previous number(ie, 6) and link 6.next to None and then connect 9.next to 1, to get the desired result. But, I am having some issues in the code it is running infinite time and a little bit wrong. Can you please take a look into this and help me? Thank you

class Node: # Class will intantiate a node object def __init__(self, data): self.data = data # node have a data self.next = None # node have a link as well ( of the next node), which is None currently.

 def typeinput(a): lis = [int(ele) for ele in a.split()] head = None tail = None count = 0 for curele in lis: if curele == -1: break newnode = Node(curele) if head is None: head = newnode tail = newnode else: tail.next = newnode tail = newnode return head

 def printll(head): while head is not None: print(str(head.data) + "->", end="") head = head.next print("None")

 def length(head): # This function is used for determining the lenght of the Linked List c = 0 while head is not None: c += 1 head = head.next return c

 def appendN(head, n): count = length(head) - n #it will give me the index of first last digit we want to break and append on first. i = 0 tail = head #we don't want to change head so, we made a tail which will keep increasing till i<count. if count < 0 or count > length(head): #if count less than 0 or greater than length print(f"Please enter the number between 0 to {length(head)}") return while i < count: #loop run until we reach previous of the first last no. we want to append tail = tail.next i += 1 headT = tail #as we find that number we assigned it to new head tail.next = None #this tail.next is of head so we make it None while headT.next is not None: # we will make loop and assign the head to the last numbers.next headT = headT.next headT.next = head return headT

 lis = "9 8 7 7 6 5 4 3 3 2 1" head = typeinput(lis) c = appendN(head, 3) printll(c)

I hope you get the Idea what I am trying to do here

Answer 1

You might want to try another implementation like the following. Note that this code works even with negative integer n values, so if you enter -3 as a parameter, the LinkedList will shift on the opposite direction, avoiding issues and exceptions:

 def appendN(head, n): if not head: return size = 1 tail = head while tail.next: size += 1 tail = tail.next tail.next = head for i in range(-n % size): tail = tail.next head = tail.next tail.next = None return head

Answer 2

You can do the operation you're attempting (which I would call a right-rotation, not an "append"), by keeping track of two node references that you advance together through the list, n nodes apart.

 def appendN(head, n): tail = new_tail = head # initialize two references into the list for _ in range(n): # advance the first one n times if not tail.next: raise ValueError("Linked list is too short") # or print and return tail = tail.next while tail.next: # then advance both in lockstep until you reach the end tail = tail.next new_tail = new_tail.next tail.next = head # link the end of the list to the beginning head = new_tail.next # grab a reference to the new head new_tail.next = None # and split the list where the second reference ended up return head