题解 | #合并k个已排序的链表#

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
#
# @param lists ListNode类一维数组
# @return ListNode类
#
class Solution:
    def merge(self, pHead1: ListNode, pHead2: ListNode) -> ListNode:
        if not pHead1:
            return pHead2
        if not pHead2:
            return pHead1
        head = ListNode(-1)
        cur = head
        while pHead1 and pHead2:
            if pHead1.val <= pHead2.val:
                cur.next = pHead1
                pHead1 = pHead1.next
            else:
                cur.next = pHead2
                pHead2 = pHead2.next
            cur = cur.next
        if pHead1:
            cur.next = pHead1
        else:
            cur.next = pHead2
        return head.next

    def divideList(self, lists: list[ListNode], left: int, right: int):
        if left > right:
            return None
        elif left == right:
            return lists[left]
        mid = int((left + right) / 2)
        return self.merge(
            self.divideList(lists, left, mid), self.divideList(lists, (mid + 1), right)
        )

    def mergeKLists(self, lists: list[ListNode]) -> ListNode:
        # write code here
        return self.divideList(lists, 0, len(lists) - 1)