Sandbox

Problem Solution # 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 # find the middle of the linked list # https://leetcode.com/problems/middle-of-the-linked-list/ slow = head fast = head.next..

Problem Solution My initial approach was to compare the head nodes from each list, put the smallest onto the result linked list, and proceed the pointer that was on the smallest node. The time complexity for this approach would be O(kN) where k is the number of elements in lists and N is the number of nodes in the final linked list. I did not find a clean way to implement this. Solution using me..