<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>

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def sortList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        split = []
        
        node = head
        while node:
            split.append([node])
            node = node.next
        
        if len(split) < 2:
            return head
        
        ordered_nodes = self.merge(split)


        for i in range(len(ordered_nodes) - 1):
            ordered_nodes[i].next = ordered_nodes[i + 1]
        ordered_nodes[-1].next = None

        return ordered_nodes[0]

    def merge(self, next_split):
        split = []
        
        while len(next_split) > 1:
            split = next_split
            next_split = []

            while split:
                left = split.pop()
                if not split:
                    next_split.append(left)
                    continue
                right = split.pop()

                new_seq = []

                while left and right:
                    if left[0].val < right[0].val:
                        new_seq.append(left.pop(0))
                    else:
                        new_seq.append(right.pop(0))
                if left:
                    new_seq.extend(left)
                elif right:
                    new_seq.extend(right)
                
                next_split.append(new_seq)
        
        return next_split[0]
    
37200