<p>Given the <code>head</code> of a singly linked list, group all the nodes with odd indices together followed by the nodes with even indices, and return <em>the reordered list</em>.</p>

<p>The <strong>first</strong> node is considered <strong>odd</strong>, and the <strong>second</strong> node is <strong>even</strong>, and so on.</p>

<p>Note that the relative order inside both the even and odd groups should remain as it was in the input.</p>

<p>You must solve the problem&nbsp;in <code>O(1)</code>&nbsp;extra space complexity and <code>O(n)</code> time complexity.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/10/oddeven-linked-list.jpg" style="width: 300px; height: 123px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,4,5]
<strong>Output:</strong> [1,3,5,2,4]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/10/oddeven2-linked-list.jpg" style="width: 500px; height: 142px;" />
<pre>
<strong>Input:</strong> head = [2,1,3,5,6,4,7]
<strong>Output:</strong> [2,3,6,7,1,5,4]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the linked list is in the range <code>[0, 10<sup>4</sup>]</code>.</li>
	<li><code>-10<sup>6</sup> &lt;= Node.val &lt;= 10<sup>6</sup></code></li>
</ul>

# 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
        odd = head
        even = head.next
        ehead = even
        while even != None and even.next != None:
            odd.next = odd.next.next
            even.next = even.next.next
            odd=odd.next
            even= even.next
        odd.next = ehead
        return head
        
18600