给你一个链表数组,每个链表都已经按升序排列。
请你将所有链表合并到一个升序链表中,返回合并后的链表。
示例 1:
输入:lists = [[1,4,5],[1,3,4],[2,6]]
输出:[1,1,2,3,4,4,5,6]
解释:链表数组如下:
[
1->4->5,
1->3->4,
2->6
]
将它们合并到一个有序链表中得到。
1->1->2->3->4->4->5->6
示例 2:
输入:lists = []
输出:[]
示例 3:
输入:lists = [[]]
输出:[]
提示:
k == lists.length
0 <= k <= 10^4
0 <= lists[i].length <= 500
-10^4 <= lists[i][j] <= 10^4
lists[i] 按 升序 排列
lists[i].length 的总和不超过 10^4
#最直观的思路
利用python里面的最小堆
# 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[ListNode]) -> ListNode:
if not lists:
return None
if len(lists) == 1:
return lists[0]
res = []
for node in lists:
while node:
heapq.heappush(res, node.val)
node = node.next
dummy = ListNode(-1)
cur = dummy
while res:
cur.next = ListNode(heapq.heappop(res))
cur = cur.next
return dummy.next