合并K个排序链表

合并K个排序链表

描述

合并 k 个排序链表，返回合并后的排序链表。请分析和描述算法的复杂度。

示例:

输入:
[
  1->4->5,
  1->3->4,
  2->6
]
输出: 1->1->2->3->4->4->5->6

题解

使用heapq，但是ListNode没有比较函数，所以需要自行定义：
ListNode.__lt__ = lambda x, y: x.val < y.val

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

ListNode.__lt__ = lambda x, y: x.val < y.val

class Solution:
    def mergeKLists(self, lists: List[ListNode]) -> ListNode:
        heap = []
        for each_list in lists:
            if each_list:
                heapq.heappush(heap, each_list)

        header = curr = ListNode(0)

        while heap:
            node = heapq.heappop(heap)
            curr.next = ListNode(node.val)
            curr = curr.next
            node = node.next
            if node:
                heapq.heappush(heap, node)

        return header.next

题解2

不定义ListNode.__lt__，在heappush的时候第一个为node.val，当val相等时，需要提供一个可比较的值，否则ListNode不可比较会报错，在此第二个数设置id，或者其他编号也可以

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def mergeKLists(self, lists: List[ListNode]) -> ListNode:
        heap = []
        for each_list in lists:
            if each_list:
                heapq.heappush(heap, (each_list.val, id(each_list), each_list))

        header = curr = ListNode(0)

        while heap:
            val, id_list, node = heapq.heappop(heap)
            curr.next = ListNode(val)
            curr = curr.next
            node = node.next
            if node:
                heapq.heappush(heap, (node.val, id_list, node))

        return header.next