重点思想： 第一轮从k个list中拿出第一个（若每个list非空的话；有可能为空）放入minheap中，minheap一直储存k个lists中最小的那个数，对顶代表当前最小的数，需要插入result listnode。所有list每个node都要加入minheap中，所以结束条件就是minheap为空，已经便利完所有node/

Time：一共kn个node，每次插入需要O(logk)，一共 O(knlogk)

Space: O(k) minheap 大小

重点：每一次将一个list中的值插入miheap中时要将指针移到下一位，方便下一次根据list的k值来加入lists[k]的下一位。

        for i in range(k):if lists[i]:#current listnode is not Noneheappush(h, [lists[i].val,i])lists[i] = lists[i].next#prepare for next around of adding into the heapcur = cur.next#append the next node of list k if list k has other nodesif lists[cur_k]:heappush(h,[lists[cur_k].val, cur_k])lists[cur_k] = lists[cur_k].nextreturn res.next

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
from heapq import *
class Solution:def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:h = []k = len(lists)for i in range(k):if lists[i]:#current listnode is not Noneheappush(h, [lists[i].val,i])lists[i] = lists[i].next#prepare for next around of adding into the heapres = ListNode(0)cur = reswhile h:#while heap still has remaining node leftcur_smallest_val, cur_k = heappop(h)cur.next = ListNode(cur_smallest_val)cur = cur.next#append the next node of list k if list k has other nodesif lists[cur_k]:heappush(h,[lists[cur_k].val, cur_k])lists[cur_k] = lists[cur_k].nextreturn res.next

来源：题目图片来源力扣（LeetCode）

著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。