LeetCode 23. Merge k Sorted Lists--Python解法--优先队列，分治法

题目地址：Merge k Sorted Lists - LeetCode

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

Example:

Input:
[1->4->5,1->3->4,2->6
]
Output: 1->1->2->3->4->4->5->6

这道题目是经典的对K个有序链表进行合并，可以使用最小堆或者说优先队列。

如果采用暴力的方法，也就是把所有节点加入数组，排序，那么时间复杂度是O(NlogN)。

采用最小堆的时间复杂度是O(NlogK)。

Python解法如下：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = Nonefrom queue import PriorityQueue
class Solution:def mergeKLists(self, lists: list) -> ListNode:dummy = ListNode(None)curr = dummyq = PriorityQueue()for node in lists:if node: q.put((node.val,id(node),node))while q.qsize()>0:curr.next = q.get()[2]curr=curr.nextif curr.next: q.put((curr.next.val,id(curr.next), curr.next))return dummy.next

如果采用分治法，时间复杂度相同。

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
from queue import PriorityQueue
class Solution:def mergeKLists(self, lists: List[ListNode]) -> ListNode:amount = len(lists)interval = 1while interval < amount:for i in range(0, amount - interval, interval * 2):lists[i] = self.merge2Lists(lists[i], lists[i + interval])interval *= 2return lists[0] if amount > 0 else listsdef merge2Lists(self, l1, l2):head = point = ListNode(0)while l1 and l2:if l1.val <= l2.val:point.next = l1l1 = l1.nextelse:point.next = l2l2 = l1l1 = point.next.nextpoint = point.nextif not l1:point.next=l2else:point.next=l1return head.next

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