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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