Odd Even Linked List(LeetCode)

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:

Given 1->2->3->4->5->NULL,
return 1->3->5->2->4->NULL

Steps:

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

class Solution(object):
    def oddEvenList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if head is None:
            return None
        odd=head
        even=head.next
        evenHead=even
        while even is not None and even.next is not None:
            odd.next=even.next
            odd=odd.next
            even.next=odd.next
            even=even.next
        odd.next=evenHead
        return head