Given the head of a linked list, return the list after sorting it in ascending order.
Example 1:
4 -> 2 -> 1 -> 3
v
1 -> 2 -> 3 -> 4
Input: head = [4,2,1,3]
Output: [1,2,3,4]
Example 2:
-1 -> 5 -> 3 -> 4 -> 0
v
-1 -> 0 -> 3 -> 4 -> 5
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)?
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
|
# 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 head.next is None:
return head
left = head
slow, fast = head, head.next
while fast and fast.next:
slow = slow.next
fast = fast.next.next
right = slow.next
slow.next = None
l1 = self.sortList(left)
l2 = self.sortList(right)
return self.mergeList(l1, l2)
def mergeList(
self,
l1: Optional[ListNode],
l2: Optional[ListNode]
) -> Optional[ListNode]:
if not l1 or not l2:
return l1 or l2
dummy = prev = ListNode()
while l1 and l2:
if l1.val > l2.val:
prev.next = l2
l2 = l2.next
else:
prev.next = l1
l1 = l1.next
prev = prev.next
prev.next = l1 or l2
return dummy.next
|