<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]:
        def mergelist(list1, list2):
            dummy=ListNode(-1)
            temp=dummy
            while(list1 is not None and list2 is not None):
                if(list1.val<list2.val):
                    temp.next=list1
                    list1=list1.next
                else:
                    temp.next=list2
                    list2=list2.next
                temp=temp.next
            if list1 is not None:
                temp.next=list1
            else:
                temp.next=list2

            return dummy.next
        def findmiddle(head):
            slow=head
            fast=head.next
            while(fast and fast.next):
                slow=slow.next
                fast=fast.next.next
            return slow


        if not head or not head.next:
            return head
        
        middle=findmiddle(head) 

        right=middle.next
        middle.next=None
        left=head

        left=self.sortList(left)
        right=self.sortList(right)

        return mergelist(left,right)    
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