23. Merge K Sorted Lists

Hard

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.

 # Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

from queue import PriorityQueue

class Solution:
    def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:

        class Wrapper():
            def __init__(self, node):
                self.node = node
            def __lt__(self, other):
                return self.node.val < other.node.val


        q = PriorityQueue()

        for l in lists:
            if l:
                q.put(Wrapper(l))

        senti = ListNode(0)
        head = senti
        while not q.empty():
            node = q.get().node
            head.next = node
            head = head.next
            if node.next:
                q.put(Wrapper(node.next))

        return senti.next



#         heap = []
#         heapq.heapify(heap)

#         for l in lists:

#             head = l
#             while head:
#                 heapq.heappush(heap, head.val)
#                 head = head.next

#         senti = ListNode(0)
#         head = senti
#         while heap:
#             head.next = ListNode(heapq.heappop(heap))
#             head = head.next

#         return senti.next

23. Merge K Sorted Lists

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