"""
Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.

You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.

Example 1:

Input: 1->2->3->4->5->NULL
Output: 1->3->5->2->4->NULL
Example 2:

Input: 2->1->3->5->6->4->7->NULL
Output: 2->3->6->7->1->5->4->NULL
Note:

The relative order inside both the even and odd groups should remain as it was in the input.
The first node is considered odd, the second node even and so on ...
"""

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def oddEvenList(self, head: ListNode) -> ListNode:
        if not head:
            return None

        odd_current = head
        even_current = head.next
        even_head = head.next
        current = head
        i = 0
        while current:
            if i % 2 == 0:
                even_current.next = current
                even_current = even_current.next
            else:
                odd_current.next = current
                odd_current = odd_current.next

            i += 1
            current = current.next




        return head

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