方法一

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:tempArray = list()for l in lists:while l:tempArray.append(l.val)l = l.nexttempArray.sort(reverse = False)resNode = tempNode = ListNode(0)for t in tempArray:tempNode.next = ListNode(t)tempNode = tempNode.nextreturn resNode.next

方法二

#还有一个类似方法一的做法，就是通过python的一个库-->heapq来实现push和pop操作，其中弹出的元素为堆中最小的element#heappush(heap, item) # pushes a new item on the heap#item = heappop(heap) # pops the smallest item from the heapclass Solution(object):def mergeKLists(self, lists):""":type lists: List[ListNode]:rtype: ListNode"""if not lists or len(lists) == 0:return Noneimport heapqheap = []# 首先 for 嵌套 while 就是将所有元素都取出放入堆中for node in lists:while node:heapq.heappush(heap, node.val)node = node.nextdummy = ListNode(None)cur = dummy# 依次将堆中的元素取出(因为是小顶堆，所以每次出来的都是目前堆中值最小的元素），然后重新构建一个列表返回while heap:temp_node = ListNode(heappop(heap))cur.next = temp_nodecur = temp_node#print(heap)return dummy.next