/** * 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…

http://www.geeksforgeeks.org/merge-sort-for-linked-list/ #include <iostream> #include <vector> #include <algorithm> #include <queue> #include <stack> #include <string> #include <fstream> #include <map> #incl…

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 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…

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;…

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   /** * Definition for singly-linked lis…

题目:Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 解析:合并k个已经有序的单链表,使其最终成为一个有序的单链表.原理就是归并排序,递归运算.基本算法recusion 与 merge 编码: public ListNode mergeKLists(ListNode[] lists) { if(lists == null || lists.leng…

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.…

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] 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…