Leetcode 23. Merge K sorted lists

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

 

思路：参考：http://www.cnblogs.com/zuoyuan/p/3772372.html

一开始的思路类似merge two sorted lists。比较所有list head.val，那个最小就把dummy指向那个ListNode，然后把相应的那个list的head往后挪一格。但是由于K不知道是多少，比较难操作。根据提示使用heap。把所有的head都push进一个heap。pop出值最小的那个，把dummy指向那个元素。然后将对应的那一个list中的下一个元素（新的head node）pushi进heap，重复上述操作。

注意和heap相关的几个操作。heapq.heapify, heapq.pop(heap), heapq.push(heap, xxxx)

 1 class Solution(object):
 2     def mergeKLists(self, lists):
 3         """
 4         :type lists: List[ListNode]
 5         :rtype: ListNode
 6         """
 7         heap = []
 8         for node in lists:
 9             if node:
10                 heap.append((node.val, node))
11         heapq.heapify(heap)
12         dummy = ListNode(0)
13         curr = dummy
14         while heap:
15             pop = heapq.heappop(heap)
16             curr.next = ListNode(pop[0])
17             curr = curr.next
18             if pop[1].next:
19                 heapq.heappush(heap, (pop[1].next.val, pop[1].next))
20         return dummy.next
21         
22         

  用K的元素构造heap的complexity是KlogK, 然后将一个元素push进heap的complexity是logK。所以总复杂度是KlogK + nlogK

转载于:https://www.cnblogs.com/lettuan/p/6443491.html