Odd Even Linked List.py

#!/usr/bin/python

'''
Given the head of a singly linked list, group all the nodes with odd indices together followed by the nodes with even indices, and return the reordered list.
The first node is considered odd, and the second node is even, and so on.
Note that the relative order inside both the even and odd groups should remain as it was in the input.
You must solve the problem in O(1) extra space complexity and O(n) time complexity.
'''

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def oddEvenList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if head == None or head.next == None:
            return head
    
        even, odd = head, head.next
        odd_head = odd
    
        while odd != None and odd.next != None:
            even.next = even.next.next
            even = even.next
            odd.next = odd.next.next
            odd = odd.next
        even.next = odd_head
        return head