将k个有序的链表拼接成一个有序的链表。
注意点:
例子:
输入: lists = [[1->2>10],[3->9],[5->6]]
输出: 1->2->3->5->6->9->10
整体思路与Merge Two Lists相同。不过就是从原来的两个数中取最小的节点改为从k个数中取最小的节点。这是一个典型的堆排序的应用,Python中堆排序可以用heapq实现。
import heapq
# Definition for singly-linked list.
class ListNode(object):
def __init__(self, x):
self.val = x
self.next = None
class Solution(object):
def mergeKLists(self, lists):
"""
:type lists: List[ListNode]
:rtype: ListNode
"""
heap = []
for node in lists:
if node:
heapq.heappush(heap, (node.val, node))
temp = ListNode(-1)
head = temp
while heap:
smallestNode = heapq.heappop(heap)[1]
temp.next = smallestNode
temp = temp.next
if smallestNode.next:
heapq.heappush(heap, (smallestNode.next.val, smallestNode.next))
return head.next
欢迎查看我的Github来获得相关源码。
原文:http://blog.csdn.net/u013291394/article/details/50391995