题解 | #合并k个已排序的链表#

# from sys import modules
# 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]):
        # write code here
        res_head = ListNode(0)
        cur_node = res_head

        nums_list = []  # 数字列表 
        for node in lists:
            if node == None: # 判断链表节点非空
                continue
            while node.next != None: # 取数，存数
                nums_list.append(node.val)
                # node = node.next
                # print(node.val)
                node = node.next
            nums_list.append(node.val)
            # print(node.val)
        # print(lists)
        # print(nums_list)
        list_data = sorted(nums_list) # 排序 
        # print(list_data)
        for num in list_data: # 链表依次 链接
            cur_node.next = ListNode(num) # 区分与C++，每次next要定义一个新的对象链表结点。
            cur_node = cur_node.next

        # cur_node.next = None 
        # print(cur_node)
        return res_head.next