T23.合并K个升序链表

linked list, divide and conquer, merge sort, https://leetcode.cn/problems/merge-k-sorted-lists/

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

python

# 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]:
        def merge(p, q):
            dummy = ListNode()
            pre = dummy
            while p and q:
                if p.val <= q.val:
                    pre.next = p
                    pre = p
                    p = p.next
                elif p.val > q.val:
                    pre.next = q
                    pre = q
                    q = q.next

            pre.next = p if p else q
            
            return dummy.next


        if not lists:
            return None

        n = len(lists)
        while n > 1:
            new_lists = []
            for i in range(0, n, 2):
                if i < n - 1:
                    new_lists.append(merge(lists[i], lists[i+1]))
                else:
                    new_lists.append(lists[i])
            lists = new_lists
            n = len(lists)
        
        return lists[0] if lists else None

T23.合并K个升序链表 ​

T23.合并K个升序链表