148_Sort_List.py

"""
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)?

"""

# 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
        
        #split list into two halfs
        left = head
        right = self.getMid(head)
        temp = right.next
        right.next = None
        right = temp

        left = self.sortList(left)
        right = self.sortList(right)
        return self.merge(left,right)

    def getMid(self,head):
        slow , fast = head , head.next
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
        return slow
    
    def merge(self, list1 , list2):
        dummy = ListNode()
        tail = dummy
        while list1 and list2:
            if list1.val < list2.val:
                tail.next = list1
                list1 = list1.next
            else:
                tail.next = list2
                list2 = list2.next
            tail = tail.next
        if list1:
            tail.next = list1
        if list2:
            tail.next = list2
        
        return dummy.next