<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 mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:

        r = ListNode()
        copy = r

        while list1 != None and list2 != None :
            if(list1.val <= list2.val):
                copy.next = ListNode(val = list1.val)
                copy = copy.next
                list1 = list1.next
            else:
                copy.next = ListNode(val = list2.val)
                copy = copy.next
                list2 = list2.next

        while list1 != None:
            copy.next = ListNode(val = list1.val)
            copy = copy.next
            list1 = list1.next
        
        while list2 != None:
            copy.next = ListNode(val = list2.val)
            copy = copy.next
            list2 = list2.next

        return r.next

    def merge(self,listH):

        head = listH
        head2 = listH
        before = None
        while head2 != None:
            before = head
            head = head.next            
            head2 = head2.next
            if(head2 != None):
                head2 = head2.next
        
        second = before.next
        if(second == None):
            return listH
        
        before.next = None

        l = self.merge(listH)
        r = self.merge(second)

        return self.mergeTwoLists(l,r) 

        
    def sortList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if(head == None):
            return None
            
        return self.merge(head)
36900