题解 | #合并k个已排序的链表|时空O(k*n)#

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
#
# @param lists ListNode类一维数组
# @return ListNode类
#
class Solution:
    def mergeKLists(self, lists: List[ListNode]) -> ListNode:
        if len(lists) == 0:
            return None
        if len(lists) == 1:
            return lists[0]
		# 分治：将问题转换为第1个链表与k-1个链表合并的问题
        return self.merge(lists[0], self.mergeKLists(lists[1:]))

	# 合并两个有序链表
    def merge(self, list1: ListNode, list2: ListNode) -> ListNode:
        if list1 == None:
            return list2
        if list2 == None:
            return list1
        if list1.val < list2.val:
            list1.next = self.merge(list1.next, list2)
            return list1
        if list1.val >= list2.val:
            list2.next = self.merge(list2.next, list1)
            return list2