<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.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
        ListNode* middleNode(ListNode* head) {
        ListNode* fast = head;
        ListNode* slow = head;
        while(fast->next!=NULL){
            fast = fast->next;
            if(fast->next!=NULL){
                fast = fast->next;
                slow = slow->next;
            }
        }
        return slow;

    }
        ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) {
        ListNode * newHead ;
        ListNode *temp;

        if(list1 == NULL && list2 == NULL)
        return NULL;
        else if(list1 == NULL && list2!= NULL)
        return list2;
        else if(list2 == NULL && list1!=NULL)
        return list1;

        if(list1->val<list2->val){
           newHead = list1;
           list1 = list1->next;
        }
        else{
         newHead = list2;
         list2 = list2->next;
        }
         temp = newHead;
        while(list1!= NULL && list2!= NULL){
            if(list1->val < list2->val){
                temp->next = list1;
                temp = list1;
                list1 = list1->next;
                if(list1 == NULL){
                    temp ->next = list2;
                }
            }
            else{
                temp->next = list2;
                temp = list2;
                list2 = list2->next;
                if(list2 == NULL){
                    temp ->next = list1;
                }
            }
        }
        if(list1 == NULL && list2 != NULL)
        temp->next = list2;

        else if(list2 == NULL && list1 != NULL) 
        temp->next = list1;

        return newHead;
    }
    ListNode* sortList(ListNode* head) {
        if(head == NULL || head->next == NULL)
        return head;

        ListNode * middle = middleNode(head);

        ListNode* right = middle->next;
        middle->next = NULL;
        ListNode* left = head;

        left = sortList(left);
        right = sortList(right);

         return mergeTwoLists(left,right);
    }
};
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