problem: Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. Tags Divide and Conquer Linked List Heap 合并K个已序单链表 thinking: (1)题目没有要求不能够新开ListNode,所以暴力破解法:提取K个list的keyword.排序.新建结点插入.这样的情况对原list是否排好序没有要求. 排…

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…

1. Merge Two Sorted Lists 我们先来看这个 问题: 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. 是的,这是一个非常简单链表操作问题.也许你只需要花几分种便能轻松写出代码. 2. Merge k Sorted Lists 我们现在来研究这…

Merge k Sorted Lists Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 两个方法: 方法1. 利用 STL 中的 multiset (根据结点内的值)自动对指针排序.空间 O(N), 时间 O(NlogN). (亦可用于 k 个无序链表).(AC: 164ms) /** * Definition for singly-linked…

Merge k Sorted Lists Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.   采用优先队列priority_queue 把ListNode放入优先队列中,弹出最小指后,如果该ListNode有下一个元素,则把下一个元素放入到队列中     /** * Definition for singly-linked list. * stru…

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…

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…

leetcode-algorithms-23 Merge k Sorted Lists 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 解法…

堆是计算机科学中一类特殊的数据结构的统称.堆通常是一个可以被看做一棵树的数组对象.在队列中,调度程序反复提取队列中第一个作业并运行,因为实际情况中某些时间较短的任务将等待很长时间才能结束,或者某些不短小,但具有重要性的作业,同样应当具有优先权.堆即为解决此类问题设计的一种数据结构. 1 定义 n个元素序列{k1,k2...ki...kn},当且仅当满足下列关系时称之为堆:(ki <= k2i, ki <= k2i+1)或者(ki >= k2i, ki >= k2i+1), (i =…

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…

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. 1,类似于Merge Sort的方法做k-1次,每次合并两个,但是这种方法超时. for(int i = 1; i < lists.size(); i++) head = merge(head, lists[i]); 2,分治法,合并的时间复杂度O(logN),不用递归,空间复杂度O(1) 思想很简单…

21.Merge Two Sorted Lists 初始化一个指针作为开头,然后返回这个指针的next class Solution { public: ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { ListNode* dummy = ); ListNode* p = dummy; while(l1 && l2){ if(l1->val <= l2->val){ p->next = l1; p = p-&…

一.题目说明 这个题目是23. Merge k Sorted Lists,归并k个有序列表生成一个列表.难度为Hard,实际上并不难,我一次提交就对了. 二.我的解答 就是k路归并,思路很简单,实现也不难. #include<iostream> #include<vector> using namespace std; struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {}…

题目要求:Merge k Sorted Lists Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. Analysis The simplest solution is using PriorityQueue. The elements of the priority queue are ordered according to their natu…

Question 23. Merge k Sorted Lists Solution 题目大意:合并链表数组(每个链表中的元素是有序的),要求合并后的链表也是有序的 思路:遍历链表数组,每次取最小节点 Java实现: public ListNode mergeKLists(ListNode[] lists) { ListNode preHead = new ListNode(0); ListNode minNode = getMinNode(lists); ListNode tmpNode =…

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…

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 题解: 归并思想. Solution 1 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {…

Merge k sorted linked lists and return it as one sorted list. 题意:把k个已经排好序的链表整合到一个链表中,并且这个链表是排了序的. 题解:这是一道经典好题,值得仔细一说. 有两种方法,假设每个链表的平均长度是n,那么这两种方法的时间复杂度都是O(nklogk). 方法一: 基本思路是:把k个链表开头的值排个序,每次取最小的一个值放到答案链表中,这次取完之后更新这个值为它后面的一个值.接着这么取一直到全部取完.那么每次更新之后怎么对当…

题目: Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. Example Given lists: [ 2->4->null, null, -1->null ], return -1->2->4->null. 题解: Solution 1 () class Solution { public: struct compare…

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…

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…

问题: Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 官方难度: Hard 翻译: 合并k个已排序的链表,得到一个新的链表并且返回其第一个节点.分析并阐述其复杂度. 这是No.021(Merge Two Sorted Lists)的深入研究. 可以借鉴归并排序的思想,对于长度为k的数组,依次进行二路归并,返回这两个链表合并之后的头结点(利用No.…

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…

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 使用分治法,时间复杂度是nlogk, n是所有元素个数的总和,k是k个lists, 这种方法和依次merge每一个list的方法的时间复杂度不同. 如果第一个list有n-k个元素,其余每个list是1个元素, 两两分治合并,每个元素参与了logk次合并, 如果是依次合并(第一个和第二个合并,之后的结…

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…