148. Sort List

Updated: 2023-09-01
2 min read

LeetCode problem

Naive Solution

  1. Traverse the linked list, adding each node’s value to a Python list
  2. sort that list
  3. create a new linked list from the sorted values
  4. return the head of this new list.

This solution would have a time complexity of O(n log n) due to the sort operation and a space complexity of O(n) because of the extra list we’re creating.

class Solution:
    def sortList(self, head):
        values = []

        node = head
        while node:
            node = node.next


        # Create a new linked list from the sorted values
        node = head
        for val in values:
            node.val = val
            node = node.next

        return head


Using the Merge Sort algorithm.

  1. Divide and Conquer: Merge sort uses the divide and conquer strategy, where we continuously split the linked list in half until we have multiple sublists of length 1. A list of length 1 is technically always sorted.
  2. Merge Sublists: Once we have the sorted sublists, we start merging them together in a manner that the resultant list is also sorted.

The trick to making this solution O(1) space complexity is to modify the existing nodes’ next pointers to generate the sorted list, rather than creating new nodes.

class Solution:
    def sortList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if not head or not head.next:
            return head

        slow = head
        fast = head.next

        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next

        mid = slow.next
        slow.next = None # separate left and right halves of linked list

        left = self.sortList(head)
        right = self.sortList(mid)

        def merge(left, right):
            if not left or not right:
                return left or right

            if left.val > right.val: # sort
                left, right = right, left

            left.next = merge(left.next, right)
            return left

        res = merge(left, right)

        return res