23. 合并K个升序链表

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：[]
 

---

 

思路1：优先队列

T：O（n）

# 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]:
        import heapq 
        dummy = ListNode(0)
        p = dummy 
        h = []
        for i in range(len(lists)):
            if lists[i]:
                heapq.heappush(h, (lists[i].val, i))
                lists[i] = lists[i].next 
        while h:
            val, idx = heapq.heappop(h)
            p.next = ListNode(val)
            p = p.next 
            if lists[idx]:
                heapq.heappush(h, (lists[idx].val, idx))
                lists[idx] = lists[idx].next 
        return dummy.next 

思路2：分而治之

T：O（n），n为节点的个数

class Solution:
    def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:
        if not lists:
            return


        n = len(lists)
        return self.merge(lists, 0, n - 1)
        
    def merge(self, lists, left, right):
        if left == right:
            return lists[left]
        mid = left + (right - left) // 2
        l1 = self.merge(lists, left, mid)
        l2 = self.merge(lists, mid + 1, right)
        return self.helper(l1, l2)
        
    def helper(self, l1, l2):
        if not l1:
            return l2
        if not l2:
            return l1
        if l1.val < l2.val:
            l1.next = self.helper(l1.next, l2)
            return l1
        else:
            l2.next = self.helper(l2.next, l1)
            return l2