Merge k sorted lists

# 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]:
        heap = []
        heapq.heapify(heap)
        for i in range(len(lists)):
            if lists[i]:
                heapq.heappush(heap, (lists[i].val, i, lists[i]))
                lists[i] = lists[i].next
        dummy = ListNode()
        root = dummy
        while heap:
            curr = heapq.heappop(heap)
            node = curr[2]
            index = curr[1]
            root.next = node
            root = root.next
            if lists[index]:
                heapq.heappush(heap, (lists[index].val, index, lists[index]))
                lists[index] = lists[index].next

        return dummy.next

Merge k Sorted Lists

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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.