README_CN.md

23. 合并 K 个升序链表

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: list = []
输出: []

示例 3:

输入: [[]]
输出: []

提示:

• k == lists.length
• 0 <= k <= 104
• 0 <= lists[i].length <= 500
• -104 <= lists[i][j] <= 104
• lists[i] 按 升序 排列
• lists[i].length 的总和不超过 104

题解 (Python)

1. 题解

# 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[Optional[ListNode]]) -> Optional[ListNode]:
        heap = [(lists[i].val, i)
                for i in range(len(lists)) if lists[i] is not None]
        heapq.heapify(heap)
        hair = ListNode()
        curr = hair

        while len(heap) > 0:
            _, i = heapq.heappop(heap)
            curr.next = lists[i]
            curr = curr.next
            lists[i] = lists[i].next
            if lists[i] is not None:
                heapq.heappush(heap, (lists[i].val, i))

        return hair.next