题解 | #合并k个已排序的链表#

这、、、这不对吧，这今天谁要陷害我

#     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 mergeKLists(self , lists: List[ListNode]) -> ListNode:
        # write code here
        return self.divide_merge(lists, 0, len(lists)-1)
    def merge(self,pHead1,pHead2):
        if not pHead1:
            return pHead2
        if not pHead2:
            return pHead1
        
        if pHead1.val<pHead2.val:
            pHead1.next = self.merge(pHead1.next, pHead2)
            return pHead1
        else:
            pHead2.next = self.merge(pHead1, pHead2.next)
            return pHead2
        
    def divide_merge(self,lists,left,right):
        if left>right:return
        if left==right:
            return lists[left]
        else:
            mid = (left+right)//2
            return self.merge(self.divide_merge(lists, left, mid),self.divide_merge(lists, mid+1, right))

运行时间：1842ms 超过0.39% 用Python 3提交的代码 占用内存：115612KB 超过0.10%用Python 3提交的代码