LeetCode 328. Odd Even Linked List

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.

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 ...

java

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */class Solution {    public ListNode oddEvenList(ListNode head) {        if (head == null || head.next == null) {            return head;        }        int index = 0;        ListNode oddDummy = new ListNode(0);        ListNode evenDummy = new ListNode(0);        ListNode odd = oddDummy;        ListNode even = evenDummy;        while (head != null) {            if (index % 2 == 0) {                even.next = head;                even = even.next;            } else {                odd.next = head;                odd = odd.next;            }            head = head.next;            index++;        }        even.next = oddDummy.next;        odd.next = null;        return evenDummy.next;    }}

python

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def oddEvenList(self, head):        """        :type head: ListNode        :rtype: ListNode        """        if head is None or head.next is None:            return head        oddDummy, evenDummy = ListNode(0), ListNode(0)        odd, even = oddDummy, evenDummy        index = 0        while head is not None:            if index % 2 == 0:                even.next = head                even = even.next            else:                odd.next = head                odd = odd.next            head = head.next            index += 1        even.next = oddDummy.next        odd.next = None        return evenDummy.next