[Lintcode]128. Longest Consecutive Sequence/[Leetcode]23. Merge k Sorted Lists
128. Longest Consecutive Sequence / 23. Merge k Sorted Lists
- 本题难度: Hard/Medium
- Topic: Data Structure - heap(别人的代码1用的是heap)
Description
Merge k sorted linked lists and return it as one sorted list.
Analyze and describe its complexity.
Example
Example 1:
Input: [2->4->null,null,-1->null]
Output: -1->2->4->null
Example 2:
Input: [2->6->null,5->null,7->null]
Output: 2->5->6->7->null
我的代码
"""
Definition of ListNode
class ListNode(object):
def __init__(self, val, next=None):
self.val = val
self.next = next
"""
class Solution:
"""
@param l1: ListNode l1 is the head of the linked list
@param l2: ListNode l2 is the head of the linked list
@return: ListNode head of linked list
"""
def mergeKLists(self, lists):
# write your code here
l = len(lists)
if l == 0:
return None
left = 0
right = l-1
return self.merge(left,right,lists)
def merge(self,left,right,lists):
if left == right:
return lists[left]
if left == right-1:
return self.mergeTwoLists(lists[left],lists[right])
mid = left+(right-left)//2
return self.mergeTwoLists(self.merge(left,mid,lists),self.merge(mid+1,right,lists))
def mergeTwoLists(self, l1, l2):
# write your code here
if l1 is None:
return l2
if l2 is None:
return l1
res = pos = ListNode(0)
while (l1 and l2):
if l1.val < l2.val:
pos.next = l1
l1 = l1.next
else:
pos.next = l2
l2 = l2.next
pos = pos.next
if l1:
pos.next = l1
else:
pos.next = l2
return res.next
别人的代码
1.heap
"""
Definition of ListNode
class ListNode(object):
def __init__(self, val, next=None):
self.val = val
self.next = next
"""
import heapq
class Solution:
"""
@param l1: ListNode l1 is the head of the linked list
@param l2: ListNode l2 is the head of the linked list
@return: ListNode head of linked list
"""
def mergeKLists(self, lists):
if not lists:
return None
dummy = ListNode(-1)
curr = dummy
heap = []
for ll in lists:
while ll:
heapq.heappush(heap, ll.val)
ll = ll.next
while heap:
# smallest
node = ListNode(heapq.heappop(heap))
curr.next = node
curr = curr.next
return dummy.next
2. Priority Queue 但是python3跑不通。因为compare的operation不能重定义。
10-line python solution with priority queue
from Queue import PriorityQueue
class Solution(object):
def mergeKLists(self, lists):
dummy = ListNode(None)
curr = dummy
q = PriorityQueue() #构建一个优先队列，里面元素是val和node，默认根据第一个元素排序了
#将头结点放到优先队列中
for node in lists:
if node: q.put((node.val,node))
#将每个结点的下一个结点放入优先队列中。之后再取出最小的。
#当下一个最小的是原来它后面的元素时，下一次取出来的就是这个元素，否则，各个链表所有比它大的最小元素也都在优先队列中了。
while q.qsize()>0:
curr.next = q.get()[1] #curr表示当前位置，初始值是无意义的头结点
curr=curr.next
if curr.next: q.put((curr.next.val, curr.next)) # 如果取出来的点之后的点不为空，则放入优先队列中。
return dummy.next
使用了优先队列。我之前没有用过。
读python 堆和优先队列的使用
常见函数
#向队列中添加元素
Queue.put(item[, block[, timeout]])
#从队列中获取元素
Queue.get([block[, timeout]])
#队列判空
Queue.empty()
#队列大小
Queue.qsize()
思路
merge
二分法
将两个有序数列merge可参照[165. Merge Two Sorted Lists/21. Merge Two Sorted Lists(https://www.cnblogs.com/siriusli/p/10364804.html)
- 时间复杂度 O(nlog(n))
