# 148. Sort List

#### Medium

***

Given the `head` of a linked list, return *the list after sorting it in **ascending order***.

 

**Example 1:**

![](https://assets.leetcode.com/uploads/2020/09/14/sort_list_1.jpg)

```
Input: head = [4,2,1,3]
Output: [1,2,3,4]
```

**Example 2:**

![](https://assets.leetcode.com/uploads/2020/09/14/sort_list_2.jpg)

```
Input: head = [-1,5,3,4,0]
Output: [-1,0,3,4,5]
```

**Example 3:**

```
Input: head = []
Output: []
```

 

**Constraints:**

* The number of nodes in the list is in the range `[0, 5 * 104]`.
* `-105 <= Node.val <= 105`

&#x20;

**Follow up:** Can you sort the linked list in `O(n logn)` time and `O(1)` memory (i.e. constant space)?

```python
# 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]:
        if not head or not head.next:
            return head
        slow, fast = head, head
        previous = None
        while fast and fast.next:
            previous = slow
            slow = slow.next
            fast = fast.next.next 
        previous.next = None
        return self.merge(self.sortList(head), self.sortList(slow))
        
    def merge(self, left, right):
        dummy = tail = ListNode(None)
        while left and right:
            if left.val < right.val:
                tail.next, left = left, left.next
            else:
                tail.next, right = right, right.next
            tail = tail.next
        tail.next = left or right
        return dummy.next
        
```