23\] 合并 K 个升序链表 ```python from typing import List,Optional class ListNode: def __init__(self, val=0, next=None): self.val = val self.next = next # @lc code=start # Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next import heapq class Solution: def mergeKLists(self, lists: List[ListNode]) -> ListNode: if not lists: return None pq = [] dummy = ListNode(-1) p = dummy for head in lists: if head: heapq.heappush(pq,(head.val,id(head),head)) while pq: node = heapq.heappop(pq)[2] p.next = node if node.next: heapq.heappush(pq,(node.next.val,id(node.next),node.next)) # p指针不断前进 p = p.next return 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