思路

對鏈表元素進行獲取，然后進行sort()排序，最后通過空節點建立鏈表法重新建立一個有序的鏈表，返回頭節點即可。

# 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]:#list先存儲然后虛擬節點res=[]if not lists:return Nonefor list in lists:while list:res.append(list.val)list=list.nextdummy=ListNode()cur=dummyres.sort()for i in range(len(res)):p=ListNode(res[i])cur.next=p cur=p return dummy.next