I'm new to programming, so excuse the possibly stupid question.
I'm doing leetcodes and I got to the linked lists. I think I understand them okay, it's just that I don't know how to test my code/call my function(?)
Here's my code, I know it works since I uploaded it onto leetcode, but I would still like to be able to run it on my machine.
class Solution:
def middleNode(self, head: Optional[ListNode]) -> Optional[ListNode]:
slow = fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
return slow
I guess I have two different problems:
- the "Optional[ListNode]) -> Optional[ListNode]:" part
- and the actual calling of the function
Some questions before used the "typing" module functions like "List", so I would simply import them and they wouldn't be a problem. But I'm not really sure what to do here
To check my solutions, I write a short piece of code that I can put example inputs into
Solution = Solution()print(Solution.middleNode(head = [1,2,3,4,5,6]))
But isn't the "head" there, just a normal list? Do I have to create an extra function separately to create the "links". I've seen the creation of a linked list done by calling a function every time you want to add a new node. So would I use a for loop to add my example case?
CodePudding user response：
I think the problem on LeetCode is poorly worded. head = [1,2,3,4,5] is not really a head as it should only refer to the first item in the list - there seems to be a bit of hidden boilerplate code that creates a linked list from input list and an output list from output node.
Here's an example code that works similiar to the LeetCode task.
from typing import Optional
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class Solution:
def middleNode(self, head: Optional[ListNode]) -> Optional[ListNode]:
slow = fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
return slow
inp = [1,2,3,4,5,6]
next = None
for i in reversed(inp):
next = ListNode(i, next) # next points to head at the end of the loop
res = Solution().middleNode(next)
out = []
while res:
out.append(res)
res = res.next
print([o.val for o in out])
CodePudding user response：
to make your code work on your machine you have to implement a couple of things: First, for your first answer, you have to implement the class ListNode given at the top of the leetcode-page:
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
The you import "Optional" from typing:
from typing import Optional
Then you have the prerequisites for your code. You have to initialise the class, as you have mentioned. The only problem here is, that your variable has the same name as your class, what could cause trouble later. To finish, you have to call your function as you already did, with one little difference: This function has to be called with "head" as a variable of type ListNode, not List, and gives you back a variable of the type ListNode.
In a nutshell, this would be my solution (of course you can and as much ListNodes as you want):
from typing import Optional
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class Solution:
def middleNode(self, head: Optional[ListNode]) -> Optional[ListNode]:
slow = fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
return slow
# Initialise class
s = Solution()
# Defining nodes of the list, assigning always the next node.
seven = ListNode(7)
six = ListNode(6, next=seven)
five = ListNode(5, next=six)
four = ListNode(4, next=five)
three = ListNode(3, next=four)
two = ListNode(2, next=three)
one = ListNode(1, next=two)
# Calling your function (with "one" as your head node of the list)
# NOTE: As this gives you back an attribute of type ListNode, you have to access the "val" attribute of it to print out the value.
print(s.middleNode(one).val)
CodePudding user response：
well if you looking for internal boilerplate, below is code for that.
here you need to create classes for nodes, linked list and solutions,.
then with the given number, you need to create a linkedlist object.
this above part is done in leetcode by themself and this object is passed to class Solution method middleNode, where OP code run and give result. next once output is got it is compared with existing solution
# Node class for individual nodes
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
# linked list class, to create a linked list of the list nodes
class LinkedList:
def __init__(self):
self.head = None
# adding a node element to linked list
def add(self, node):
if self.head is None:
self.head = node
else:
curr = self.head
while curr.next:
curr = curr.next
curr.next = node
curr = node
# printing element of existing linked list
def print_ll(self):
curr= self.head
while curr:
print(curr.val)
curr= curr.next
# leetcode solution class
class Solution:
def middleNode(self, head) :
slow = fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
return slow
# value with which we create list nodes
values = [1,2,3,4,5,6,7,8,9,10]
ll = LinkedList() # create a linked list class object
for i in values:
node = ListNode(i) # creating a list node
ll.add(node) # adding list node to linked list
#ll.print_ll() # printing linked list
x = Solution().middleNode(ll.head) # passing linked list object to leetcode solution method and getting result
while x: # printing result
print(x.val)
x=x.next
CodePudding user response：
head is pointing to the first element in the list. You should go through the topic "Pointers in C", you'll understand more.
Python uses a lot of shorthand for declaring and defining.