题解 | #合并k个已排序的链表#

# 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
#
from functools import reduce
class Solution:
    def mergeKLists(self , lists: List[ListNode]) -> ListNode:
        n = len(lists)
        if n == 0:
            return None
        elif n == 1:
            return lists[0]

        def merge(l: ListNode, r: ListNode):
            if l is None:
                return r
            elif r is None:
                return l

            ret = ListNode(None)
            cur = ret
            while all((l is not None, r is not None)):
                if l.val < r.val:
                    cur.val = l.val
                    l = l.next
                else:
                    cur.val = r.val
                    r = r.next
                
                cur.next = ListNode(None)
                cur = cur.next

            if l is not None:
                cur.val = l.val
                cur.next = l.next
            elif r is not None:
                cur.val = r.val
                cur.next = r.next
            return ret

        return reduce(merge, lists)

这里的重点在于：如果一个链表节点数为"0"时，实际传入的变量为None，这一点在python的Type hint中并没有指出(应该使用List[Optional[ListNode]])来作为输入。因此，每次reduce时不要忘了判断一下输入的是否为None

以及，由于这个问题错了好几次，导致我最后好像也忘了加Optional了 \狗头