## Description

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 …

## Solutions

### 1. Two head

```
# Time: O(n)
# Space: O(1)
# 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 or not head.next:
return head
odd, even = ListNode(0), ListNode(0)
odd_head, even_head = odd, even
idx = 0
while head:
if idx & 1 == 0:
odd.next = head
odd = odd.next
else:
even.next = head
even = even.next
head = head.next
idx += 1
even.next = None
odd.next = even_head.next
return odd_head.next
# 71/71 cases passed (44 ms)
# Your runtime beats 51.11 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (14.6 MB)
```