<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.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
   
    ListNode* oddEvenList(ListNode* head) {
       if(head==nullptr || head->next==nullptr){
           return head;
       }
       ListNode*odd=head;
       ListNode*even=head->next;
       ListNode*evenHead=even;

       while(even!=nullptr && even->next!=nullptr){
         odd->next=even->next;
         odd=odd->next;

         even->next=odd->next;
         even=even->next;
       }
       odd->next=evenHead;

       return head;
    }
};
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