23. 合并 K 个升序链表

23. 合并 K 个升序链表 #

链接 #

23. 合并 K 个升序链表

题目 #

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

解答 #

# 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]:
        if not lists:
            return None
        
        n = len(lists)
        if n == 1:
            return lists[0]
        
        if n == 2:
            return self.mergeTwoLists(lists[0], lists[1])

        h = n >> 1
        l1 = self.mergeKLists(lists[: h])
        l2 = self.mergeKLists(lists[h:])

        return self.mergeTwoLists(l1, l2)
    
    def mergeTwoLists(self, l1: Optional[ListNode], l2: Optional[ListNode]):
        d = ListNode()
        c = d
        while l1 and l2:
            if l1.val < l2.val:
                c.next = l1
                l1 = l1.next
            else:
                c.next = l2
                l2 = l2.next

            c = c.next

        c.next = l1 if l1 else l2

        return d.next