题解 | #合并k个已排序的链表#

from operator import le
# 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:
# 特殊情况特殊处理
        if len(lists)==0:
            return None
        if len(lists)==1:
            return lists[0]
        mid=len(lists)//2
        left =self.mergeKLists(lists[:mid])
        right=self.mergeKLists(lists[mid:])
## 下面就是合并两个有序链表的代码了。
        dummy=ListNode(-1)
        curr=dummy
        while left and right:
            if left.val < right.val:
                curr.next=left
                left=left.next
            else:
                curr.next=right
                right=right.next
            curr=curr.next
        if left==None:
            curr.next = right
        else:
            curr.next = left
        curr = dummy.next
        return dummy.next