题目

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

题解

1. 优先级队列

时间复杂度O(n*logk)

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def mergeKLists(self, lists: List[ListNode]) -> ListNode:
        import heapq
        dummy = ListNode(0)
        p = dummy
        head = []
        for i in range(len(lists)):
            if lists[i]:
                heapq.heappush(head, (lists[i].val, i))
                lists[i] = lists[i].next
        while head:
            val, idx = heapq.heappop(head)
            p.next = ListNode(val)
            p = p.next
            if lists[idx]:
                heapq.heappush(head, (lists[idx].val, idx))
                lists[idx] = lists[idx].next
        return dummy.next

2. 两两合并

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def mergeKLists(self, lists: List[ListNode]) -> 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.mergeTwoLists(l1, l2)
    def mergeTwoLists(self,l1, l2):
        if not l1:return l2
        if not l2:return l1
        if l1.val < l2.val:
            l1.next = self.mergeTwoLists(l1.next, l2)
            return l1
        else:
            l2.next = self.mergeTwoLists(l1, l2.next)
            return l2