<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]:
        self.node_map = {}
        i = 0
        curr = head
        while curr:
            self.node_map[i] = curr
            curr = curr.next
            i += 1
        return self.__sortList(head, 0, i)
    
    def __sortList(self, head, i, j):
        if not head or not head.next:
            return head
        left, right = self.__split(i, j)
        left = self.sortList(left)
        right = self.sortList(right)
        merged = ListNode()
        curr = merged
        while left and right:
            if left.val < right.val:
                curr.next = left
                left = left.next
            else:
                curr.next = right
                right = right.next
            curr = curr.next
        if left:
            curr.next = left
        elif right:
            curr.next = right
        return merged.next
        

    def __split(self, i, j):
        left = self.node_map[i]
        m = (i + j + 1) // 2
        right = self.node_map[m]
        self.node_map[m-1].next = None
        return left, right
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