Description
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
Python3
# 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
def merge(head1, head2):
curr = res = ListNode(-1)
while head1 and head2:
if head1.val < head2.val:
curr.next = ListNode(head1.val)
head1 = head1.next
else:
curr.next = ListNode(head2.val)
head2 = head2.next
curr = curr.next
curr.next = head1 or head2
return res.next
pre, slow, fast = None, head, head
while fast and fast.next:
pre, slow, fast = slow, slow.next, fast.next.next
pre.next = None
return merge(self.sortList(head), self.sortList(slow))