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
#
class Solution:
    def mergeKLists(self , lists: List[ListNode]) -> ListNode:
        lists = [node for node in lists if node is not None] #先把none过滤掉
        if len(lists) == 0:
            return None
        res = ListNode(-1) # 添加个临时表头
        cur = res
        while True: # 比较每个头，把最小的接到res的后面，并且把该头往后移一位。如果移动后是none，则删除
            if len(lists) == 1:
                cur.next = lists[0]
                break
            tmp = []
            for i in range(len(lists)):
                tmp.append(lists[i].val)
            min_value = min(tmp)
            min_index = tmp.index(min_value)
            cur.next = lists[min_index]
            cur = cur.next
            lists[min_index] = lists[min_index].next
            if lists[min_index] == None:
                del lists[min_index]
        return res.next