<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>

class Solution {
public:
    ListNode* merge(ListNode* left, ListNode* right)
    {
        if (!left) return right;
        if (!right) return left;
        
        ListNode root;
        ListNode* head = &root;
        auto process = [&](ListNode ** curr) {
              head->next = *curr;
              *curr = (*curr)->next;
              head = head->next;
              head->next = nullptr;
        };
        while(left && right) {
            if (left->val < right->val) {
                process(&left);
            }
            else
            {
                process(&right);
            }
        }
        if (left) head->next = left;
        if (right) head->next = right;

        return root.next;
    }

    ListNode* sortList(ListNode* head) {
        
        // null or only one elements return directly
        if (!head)
          return nullptr;
        if(!head->next)
          return head;
          
        ListNode* slow = head;
        ListNode* prev = slow;
        ListNode* fast = head;

        while (fast) {
            prev = slow;
            slow = slow->next;
            fast = fast->next;
            if (!fast)
              break;
            else
              fast = fast->next;
        }
        prev->next = nullptr;
        ListNode* left = sortList(head);
        ListNode* right = sortList(slow);
        ListNode* root = merge(left, right);
        return root;
    }
};
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