23 Merge k Sorted Lists

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 2:

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

Constraints:

  • k == lists.length
  • 0 <= k <= 10^4
  • 0 <= lists[i].length <= 500
  • -10^4 <= lists[i][j] <= 10^4
  • lists[i] is sorted in ascending order.
  • The sum of lists[i].length won’t exceed 10^4.
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# 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]:
        n = len(lists)
        if n == 0:
            return None
        elif n == 1:
            return lists[0]
        else:
            left = self.mergeKLists(lists[:n//2])
            right = self.mergeKLists(lists[n//2:])
            return self.merge2Lists(left, right)
        
    def merge2Lists(self, l1, l2):
        head = point = ListNode(0)
        while l1 and l2:
            if l1.val <= l2.val:
                point.next = l1
                l1 = l1.next
            else:
                point.next = l2
                l2 = l1
                l1 = point.next.next
            point = point.next
        if not l1:
            point.next=l2
        else:
            point.next=l1
        return head.next