Middle of the Linked List Python

Given a non-empty, singly linked list with head node head, return a middle node of linked list.

If there are two middle nodes, return the second middle node.

Example 1:

Input: [1,2,3,4,5]
Output: Node 3 from this list (Serialization: [3,4,5])
The returned node has value 3.  (The judge's serialization of this node is [3,4,5]).
Note that we returned a ListNode object ans, such that:
ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.

Example 2:

Input: [1,2,3,4,5,6]
Output: Node 4 from this list (Serialization: [4,5,6])
Since the list has two middle nodes with values 3 and 4, we return the second one.

Note:

• The number of nodes in the given list will be between 1 and 100.

# Definition for singly-linked list.

# class ListNode:

#     def __init__(self, x):

#         self.val = x

#         self.next = None

class Solution:

    def middleNode(self, head: ListNode) -> ListNode:

        pA = head

        if head is None:

            return head

        pA = head

        lA = 0

        while pA:

            lA+=1

            pA =pA.next

        pB = head

        lA = lA//2

        while lA != 0:

            pB = pB.next

            lA -= 1

        return pB if lA % 2 ==0 else pB.next

            

Time Complexity : O(n)

Space Complexity : O(1)

Fast Pointer and Slow Pointer :

# Definition for singly-linked list.

# class ListNode:

#     def __init__(self, x):

#         self.val = x

#         self.next = None

class Solution:

    def middleNode(self, head: ListNode) -> ListNode:

        slow = fast = head

        while fast and fast.next:

            slow = slow.next

            fast = fast.next.next

        return slow

    

Time Complexity : O(n)

Space Complexity : O(1)