Sort List

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

Example 1:

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

Example 2:

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

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

SOLUTION:

# 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, list2):
        if not list1 or not list2:
            return list1 or list2
        if list1.val < list2.val:
            list1.next = self.mergeTwoLists(list1.next, list2)
            return list1
        else:
            list2.next = self.mergeTwoLists(list1, list2.next)
            return list2

    def sortList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if not head or not head.next:
            return head
        prev = None
        slow = head
        fast = head
        while slow and fast and fast.next:
            prev = slow
            slow = slow.next
            fast = fast.next.next
        prev.next = None
        return self.mergeTwoLists(self.sortList(head), self.sortList(slow))