public ListNode mergeKLists(ListNode[] lists) { if(lists==null||lists.length==0) return null; PriorityQueue<ListNode> minHeap=new PriorityQueue<>((o1, o1)->{ return o1.val-o2.val; }); for(int i=0; i<lists.length; i++){ if(lists[i]!=null)…

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…

一直直到bug-free.不能错任何一点. 思路不清晰:刷两天. 做错了,刷一天. 直到bug-free.高亮,标红. 185,OA(YAMAXUN)--- (1) findFirstDuplicate string in a list of string. import java.util.HashSet; import java.util.Set; public class Solution { public static void main(String[] args) { String[…

--------------------------------------------------------------- 本文使用方法:所有题目,只需要把标题输入lintcode就能找到.主要是简单的剖析思路以及不能bug-free的具体细节原因. ---------------------------------------------------------------- ------------------------------------------- 第九周:图和搜索. ---…

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.…

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…

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…

Swap Nodes in Pairs 思路:需要构造一个头指针,指向链表.一次比较两个指针,将其翻转,临界条件是pre != null(链表长度为偶数) && pre.next != null(链表长度为奇数),然后更新头指针和pre public ListNode swapPairs(ListNode head) { if(head == null) return null; ListNode newNode = new ListNode(0); ListNode node = new…

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…

/* ** 算法的思路: ** 1.将k个链表的首元素进行建堆 ** 2.从堆中取出最小的元素,放到链表中 ** 3.如果取出元素的有后续的元素,则放入堆中,若没有则转步骤2,直到堆为空 */ #include <stdio.h> struct ListNode { int val; struct ListNode *next; }; #define PARENT(i) (((i)-1)/2) #define LEFT(i) ((i)*2+1) #define RIGHT(i) ((i)*2+…

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…

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…

数据流的中位数 class MedianFinder { private Queue<Integer> maxHeap = new PriorityQueue(new Comparator<Integer>(){ @Override public int compare(Integer i1, Integer i2){ return Integer.compare(i2, i1); } }); private Queue<Integer> minHeap = new P…

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…

/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* mergeKLists(vector<ListNode*>& lists) { const int len = lists.size();…

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…

博客搬至blog.csgrandeur.com,cnblogs不再更新. 新的题解会更新在新博客:http://blog.csgrandeur.com/2014/01/15/LeetCode-OJ-Solution/ ———————————————————————————————————————— ———————————————————————————————————————— LeetCode OJ 题解 LeetCode OJ is a platform for preparing tech…

题目: Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity 思路1: 依次归并排序,首先归并前两个,然后归并完成的链表依次和剩下的链表进行归并排序 时间复杂度为O(m*n) 代码: public static ListNode mergeKLists1(ListNode[] lists){ int len = lists.length; if(len =…

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. Summary:  Finds the smallest node every round, then links it to the one sorted list. ListNode *mergeKLists(vector<ListNode *> &lists) { ListNode *…

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. ======= 合并k个链表形成一个已排序链表 思路: 如何合并两个有序链表?经典merge算法: ListNode *mergeList(ListNode *head1,ListNode *head2){ ListNode dummy(-); ListNode *h = &dummy; while(…

题目: Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 代码: /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ cla…

题意: Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. (Hard) 分析: 方法1: 利用做过的merge 2 sorted list,将头两个归并,结果再与下一个归并,以此类推,归并完所有. 时间复杂度分析是个小问题, merge 2 sorted list的复杂度是O(n),本以为结果就是O(n * k). 但是仔细考虑,随着归并不断进行,其…

Title: Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 排好序的,然后merge,很容易让人联想到归并排序.可以将k个序列按照二分进行分割,然后当长度为1时返回一个排好序的序列,最后将两个序列merge class Solution{ public: ListNode* merge(ListNode* list1, ListNode* lis…

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. [解题思路] 以前的解法的时间复杂度过高,通过在网上搜索,得到优化的时间复杂度:O(n*lgk) 维护一个大小为k的最小堆,每次得到一个最小值,重复n次 /** * Definition for singly-linked list. * public class ListNode { * int v…

题目描述: Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 解题思路: 分治方法.将K个List不断地分解为前半部分和后半部分.分别进行两个List的合并.最后将合并的结果合并起来. 代码如下: /** * Definition for singly-linked list. * public class ListNode { * int val;…

/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ struct helper { ListNode *head; int len; helper(ListNode *h, int l) : head ( h ), len ( l ) {} }; class helper…

/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public ListNode mergeKLists(List<ListNode> lists) { ListNode head=new Lis…

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