<p>You are given the head of a singly linked-list. The list can be represented as:</p>

<pre>
L<sub>0</sub> &rarr; L<sub>1</sub> &rarr; &hellip; &rarr; L<sub>n - 1</sub> &rarr; L<sub>n</sub>
</pre>

<p><em>Reorder the list to be on the following form:</em></p>

<pre>
L<sub>0</sub> &rarr; L<sub>n</sub> &rarr; L<sub>1</sub> &rarr; L<sub>n - 1</sub> &rarr; L<sub>2</sub> &rarr; L<sub>n - 2</sub> &rarr; &hellip;
</pre>

<p>You may not modify the values in the list&#39;s nodes. Only nodes themselves may be changed.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/04/reorder1linked-list.jpg" style="width: 422px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,4]
<strong>Output:</strong> [1,4,2,3]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/09/reorder2-linked-list.jpg" style="width: 542px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,4,5]
<strong>Output:</strong> [1,5,2,4,3]
</pre>

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

<ul>
	<li>The number of nodes in the list is in the range <code>[1, 5 * 10<sup>4</sup>]</code>.</li>
	<li><code>1 &lt;= Node.val &lt;= 1000</code></li>
</ul>

#define opt() ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
class Solution {
public:
    void reorderList(ListNode* head) {
        opt();
        if(head == NULL || head->next == NULL) return;

        ListNode * slow = head;
        ListNode * firstList = head; // TO MERGE THE FIRST HALF AND REVERSED SECOND HALF
        ListNode * fast = head;
        ListNode * prevTail = NULL; // THIS IS THE LAST NODE IN THE FIRST HALF

        while(fast && fast->next){
            prevTail = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        prevTail->next = NULL;
        ListNode *secondList = reverse(slow);  // REVERSE THE SECOND HALF OF LL
        merge(firstList,secondList); // MERGE BOTH HALVES
    }
    ListNode *reverse(ListNode *head){ // BASIC REVERSAL OF LL
        ListNode *prev = NULL;
        ListNode *curr = head, *nextNode = NULL;

        while(curr != NULL){
            nextNode = curr->next;
            curr->next = prev;
            prev = curr;
            curr = nextNode;
        }
        return prev;
    }
    void merge(ListNode *l1, ListNode *l2){
        while(l2 != NULL){ // L1 || L2 != NULL ARE THE SAME THING HERE AS LIST IS SPLIT FROM MIDDLE
            ListNode *ptr1 = l1->next, *ptr2 = l2->next;
            l1->next = l2;

            if(ptr1 == NULL) break;

            l2->next = ptr1;
            l1 = ptr1;
            l2 = ptr2;
        }
    }
};
22900