<p>Given the <code>head</code> of a linked list, return <em>the list after sorting it in <strong>ascending order</strong></em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/14/sort_list_1.jpg" style="width: 450px; height: 194px;" />
<pre>
<strong>Input:</strong> head = [4,2,1,3]
<strong>Output:</strong> [1,2,3,4]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/14/sort_list_2.jpg" style="width: 550px; height: 184px;" />
<pre>
<strong>Input:</strong> head = [-1,5,3,4,0]
<strong>Output:</strong> [-1,0,3,4,5]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> head = []
<strong>Output:</strong> []
</pre>

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

<ul>
	<li>The number of nodes in the list is in the range <code>[0, 5 * 10<sup>4</sup>]</code>.</li>
	<li><code>-10<sup>5</sup> &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Can you sort the linked list in <code>O(n logn)</code> time and <code>O(1)</code> memory (i.e. constant space)?</p>

class Solution:
    def sortList(self, head: Optional[ListNode]) -> Optional[ListNode]:

        # Split the list into two halves
        def split_list(head):
            slow, fast = head, head
            while fast and fast.next:
                prev = slow
                slow = slow.next
                fast = fast.next.next
            prev.next = None  # Split the list into two halves
            return head, slow

        # Merge two sorted linked lists
        def merge_sorted_lists(l1, l2):
            dummy = ListNode()
            cur = dummy

            while l1 and l2:
                if l1.val < l2.val:
                    cur.next = l1
                    l1 = l1.next
                else:
                    cur.next = l2
                    l2 = l2.next
                cur = cur.next

            cur.next = l1 if l1 else l2
            return dummy.next

        # Recursive merge sort implementation
        def merge_sort(head):
            if not head or not head.next:
                return head

            # Split list into two halves
            left, right = split_list(head)

            # Recursively sort both halves
            left = merge_sort(left)
            right = merge_sort(right)

            # Merge the two sorted halves
            return merge_sorted_lists(left, right)

        # Start the merge sort
        return merge_sort(head)

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