Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 这道题让我们合并k个有序链表,之前我们做过一道Merge Two Sorted Lists 混合插入有序链表,是混合插入两个有序链表.这道题增加了难度,变成合并k个有序链表了,但是不管合并几个,基本还是要两两合并.那么我们首先考虑的方法是能不能利用之前那道题的解法来解答此题.答案是肯定的,但是需要修改…

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 这道题让我们合并k个有序链表,最终合并出来的结果也必须是有序的,之前做过一道 M…

作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 个人公众号:负雪明烛 本文关键词:合并,链表,单链表,题解,leetcode, 力扣,Python, C++, Java 题目地址: https://leetcode.com/problems/merge-k-sorted-lists/description/ 题目描述: Merge k sorted linked lists and return it as one sorted li…

题目:Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 题意:把k个排序成一个有序链表. 用优先队列先把k个链表遍历一遍把值存起来,在建一个新链表吧数从优先队列里一个个放进去,注意空指针的判断. /** * Definition for singly-linked list. * struct ListNode { * int val; * List…

1. Merge Two Sorted Lists 题目链接 题目要求:  Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. 这道题目题意是要将两个有序的链表合并为一个有序链表.为了编程方便,在程序中引入dummy节点.具体程序如下: /** * Definitio…

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. Input: [ 1->4->5, 1->3->4, 2->6 ] Output: 1->1->2->3->4->4->5->6 题意: 归并k个有序链表. 思路: 用一个最小堆minHeap,将所有链表的头结点放入 新建一个linkedl…

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 思路 这道题最简单的方法就是我们将K个链表中的所有数据都存进数组中,然后对数据进行…

题意 Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 将K个排好序的链表合并成一个 解法 和之前合并两个有序链表那题不同,那题用的是两个指针来分别记录两个链表中的位置,将小的那个插入新链表然后指针右移,有点像归并排序时用到的方法.这题如果用K个指针来记录的话,每次需要找出最小的那个值,比较麻烦,所以采用的优先队列,首先将所有链表的第一个值入队,然后…

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 类似于归并2个链表,暴力一点的方法就是,每取出一个list就与以前的list归并返回merge后list,知道所有list merge完成. 但是可惜,这样做会TLE.贴下代码先: /** * Definition for singly-linked list. * struct ListNode {…

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 要合并K个排好序的链表,我用的方法是用一个优先队列每次存K个元素在队列中,根据优…

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 思路:这题最容易想到的是,(假设有k个链表)链表1.2合并,然后其结果12和3合并,以此类推,最后是123--k-1和k合并.至于两链表合并的过程见merge two sorted lists的分析.复杂度的分析见JustDoIT的博客.算法复杂度:假设每个链表的平均长度是n,则1.2合并,遍历2n个…

题目描述 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 解题思路 利用分治的思想,划分k个排序链表为两半,递归的合并两部分中的链表.对于单个链表直接返回,对于两个链表调用merge函数,返回合并好的排序链表. 代码 /** * Definition for singly-li…

一天一道LeetCode系列 (一)题目 Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. (二)解题 合并K个已拍好序的链表.剑指上有合并两个已排好序的链表的算法,那么K个数,我们可以采用归并排序的思想,不过合并函数可能需要修改一下,换成合并两个已排好序的链表的方法.代码如下: /** * Definition for singly-linked…

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 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:…

转载:https://leetcode.windliang.cc/leetCode-23-Merge-k-Sorted-Lists.html 描述 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…

题目 Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 翻译 合并k个有序的链表 Hints Related Topics: LinkedList, Divide and Conquer, Heap Solution1:Divide and Conquer 参考 归并排序-维基百科 思路:分治,先分成两个子任务,然后递归子任务,最后回溯回来..这里就…

题目链接 Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 合并k个有序的链表,我们假设每个链表的平均长度是n.这一题需要用到合并两个有序的链表子过程 算法1: 最傻的做法就是先1.2合并,12结果和3合并,123结果和4合并,…,123..k-1结果和k合并,我们计算一下复杂度. 1.2合并,遍历2n个节点 12结果和3合并,遍历3n个节点 123…

1 题目 Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 2 思路 当时看到这个题目就想到的是归并排序.好吧,但是,具体代码写不出来.题目给的是数组:public ListNode mergeKLists(ListNode[] lists),我就想,怎么归并排序,然后返回一个已经排好的,再进行递归.想破脑袋没想出来. 原来,可以赋值给一个Array…

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 解题思路: 1.先取出k个list的首元素,每个首元素是对应list中的最小元素,组成一个具有k个结点的最小堆:o(k*logk) 2.此时堆顶元素就是所有k个list的最小元素,将其pop出,并用此最小元素所在list上的下一个结点(如果存在)填充堆顶,并执行下滤操作,重新构建堆.o(logk) 3…

Discription: Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. Subscribe to see which companies asked this question. 思路:其实就是归并排序的最后一步归并操作.思想是递归分治,先把一个大问题分成2个子问题,然后对2个子问题的解进行合并.经过一次遍历就能找出已经有序的序列.就算是题目中给…

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 解题思路一: 之前我们有mergeTwoLists(ListNode l1, ListNode l2)方法,直接调用的话,需要k-1次调用,每次调用都需要产生一个ListNode[],空间开销很大.如果采用分治的思想,对相邻的两个ListNode进行mergeTwoLists,每次将规模减少一半,直到…

Merge k Sorted Lists Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 题目意思: 合并K条已经排序的链表.分析时间复杂度. 解题思路: 很容易就想起之前学的合并两条链表的算法,这一题其实就是那个题目的扩展,变成合并K条了.我采用的方法就是迭代法. 如果只有一条,直接返回.如果只有两条,就只需要调用mergeTwo一下.如果超过两…

# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 23: Merge k Sorted Listshttps://oj.leetcode.com/problems/merge-k-sorted-lists/ Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. ===Comments…

https://oj.leetcode.com/problems/merge-k-sorted-lists/ 归并K已经整理阵列,和分析算法的复杂. 解决报告:无论是不考虑优化,最简单的实现是要重新走路List<ListNode>.对当中每一个链表同当前链表做一遍类似于归并排序最后一步的merge操作. 算法复杂度是O(KN) public class Solution { ListNode mergeTwoLists(ListNode list1, ListNode list2) { Lis…

Merge k Sorted Lists Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. Show Tags 参考资料: http://blog.csdn.net/linhuanmars/article/details/19899259. SOLUTION 1: 使用分治法.左右分别递归调用Merge K sorted List,然后再使用merg…

Merge Two Sorted Lists Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. 依次拼接 复杂度 时间 O(N) 空间 O(1) 思路 该题就是简单的把两个链表的节点拼接起来,我们可以用一个Dummy头,将比较过后的节点接在这个Dummy头之后.最后…

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 解法1: 采用递归的方法,不管合并几个,归根到底还是需要两两合并. 首先想到的是前两个先合并,然后再跟第三个合并,然后第四个....但是这种做法效率不高. 换个思路,采用分治法,对数量超过2的任务进行拆分,直到最后只有一个或两个链表再进行合并.代码如下: /** * Definition for si…

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 题意:对k个有序的链表进行归并排序.并分析其复杂度. /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(N…

题目:合并两个已排序链表 难度:Easy 题目内容: Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. 翻译: 合并两个已排序的链表,并将其作为一个新链表返回.新的链表应该通过将前两个列表的节点拼接在一起. Example: Input: 1->2->4, 1-&…

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 题解:最开始用了最naive的方法,每次在k个链表头中找出最小的元素,插入到新链表中.结果果断TLE了. 分析一下,如果这样做,每取出一个节点,要遍历k个链表一次,假设k个链表一共有n个节点,那么就需要O(nk)的时间复杂度. 参考网上的代码,找到了用最小堆的方法.维护一个大小为k的最小堆,存放当前k…