reorder_odd_even_linked_list.cpp

Given the head of a singly linked list, group all the nodes with odd indices together
followed by the nodes with even indices, and return the reordered list.

The first node is considered odd, and the second node is even, and so on.
Note that the relative order inside both the even and odd groups should
remain as it was in the input.
You must solve the problem in O(1) extra space complexity and O(n) time complexity.

Example 1:
Input: head = [1,2,3,4,5]
Output: [1,3,5,2,4]

Example 2:
Input: head = [2,1,3,5,6,4,7]
Output: [2,3,6,7,1,5,4]

Constraints:
The number of nodes in the linked list is in the range [0, 10^4].
-10^6 <= Node.val <= 10^6
********************************************************************************










# Approach 1:

/**
 * 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) return head;

        ListNode *odd = head, *even = head->next, *evenHead = head->next;

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

        odd->next = evenHead; // join odd and even list

        return head;
    }
};

TC -> O(n), n is the length of the linked list
SC -> O(1)

Extra:
	Here, for odd->next = evenHead; // join odd and even list
	We cannot do odd->next = head->next as the link head->next is already updated.