[23] 合并 K 个升序链表

from typing import List,Optional
class ListNode:def __init__(self, val=0, next=None):self.val = valself.next = next# @lc code=start
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = nextimport heapqclass Solution:def mergeKLists(self, lists: List[ListNode]) -> ListNode:if not lists:return Nonepq = []dummy = ListNode(-1)p = dummyfor head in lists:if head:heapq.heappush(pq,(head.val,id(head),head))while pq:node = heapq.heappop(pq)[2]p.next = nodeif node.next:heapq.heappush(pq,(node.next.val,id(node.next),node.next))# p指针不断前进p = p.nextreturn dummy.next

注意点1
对 head是否为None的判断必须有：

    for head in lists:if head:heapq.heappush(pq,(head.val,id(head),head

否则过不了测试用了[None]
while pq:
p.next = node
或者 p.next = ListNode(node.val) 无区别
因为p.next 会被覆盖成小顶堆的最小值，知道没值了，指向None