leetcode【困难】合并k个升序链表

思路一：链表逐个两两合并（超时了）
参考：leetcode【简单】21、合并两个有序链表

# 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[ListNode]) -> ListNode:
        def mergeTwoLists(l1,l2):
            if not l1:
                return l2
            elif not l2:
                return l1
            elif l1.val<l2.val:
                l1.next=mergeTwoLists(l1.next,l2)
                return l1
            else:
                l2.next=mergeTwoLists(l1,l2.next)
                return l2

        res=[]
        if not lists:
            return None
        n = len(lists)
        res = lists[0]
        for i in range(1,n):
            res=mergeTwoLists(lists[i],res)
        return res

思路二：分治/二分法

# 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[ListNode]) -> ListNode:
        def mergeTwoLists(l1,l2):
            if not l1:
                return l2
            elif not l2:
                return l1
            elif l1.val<l2.val:
                l1.next=mergeTwoLists(l1.next,l2)
                return l1
            else:
                l2.next=mergeTwoLists(l1,l2.next)
                return l2
        
        def helper(begin,end):
            if begin==end:
                return lists[begin]
            mid = begin+(end-begin)//2
            left = helper(begin,mid)
            right = helper(mid+1,end)
            return mergeTwoLists(left,right)

        if not lists:
            return None

        return helper(0,len(lists)-1)