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23. Merge k Sorted Lists

Difficulty Topics

Description

You are given an array of k linked-lists lists, each linked-list is sorted in ascending order.

Merge all the linked-lists into one sorted linked-list and return it.

 

Example 1:

Input: lists = [[1,4,5],[1,3,4],[2,6]]
Output: [1,1,2,3,4,4,5,6]
Explanation: The linked-lists are:
[
  1->4->5,
  1->3->4,
  2->6
]
merging them into one sorted list:
1->1->2->3->4->4->5->6

Example 2:

Input: lists = []
Output: []

Example 3:

Input: lists = [[]]
Output: []

 

Constraints:

  • k == lists.length
  • 0 <= k <= 104
  • 0 <= lists[i].length <= 500
  • -104 <= lists[i][j] <= 104
  • lists[i] is sorted in ascending order.
  • The sum of lists[i].length will not exceed 104.

Solution

merge-k-sorted-lists.py
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:
        if not lists: return None
        n = len(lists)

        def mergeLists(start, end):
            if start == end: 
                return lists[start]
            elif start < end:
                mid = start + (end - start) // 2

                l1 = mergeLists(start, mid)
                l2 = mergeLists(mid + 1, end)

                return merge(l1, l2)
            else:
                return None

        def merge(l1, l2):
            if not l1: return l2
            if not l2: return l1

            res = curr = ListNode()

            while l1 and l2:
                if l1.val <= l2.val:
                    curr.next = ListNode(l1.val)
                    l1 = l1.next
                else:
                    curr.next = ListNode(l2.val)
                    l2 = l2.next

                curr = curr.next

            curr.next = l1 or l2

            return res.next

        return mergeLists(0, n - 1)