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))
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