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
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. 题目标签:Linked List 题目给了我们两个lists,让我们有序的合并两个 lists. 这题利用递归可以从list 的最后开始向前链接nodes,代码很简洁,清楚. Java Solution: Runti
①英文题目 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. Example: Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4->4 ②中文题目 将两个有序链表合并为一个新的有序
题目 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. Example: Input:1->2->4, 1->3->4 Output:1->1->2->3->4->4 思路 思路一:建立一个新链表 建立一个新的链表,用
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode mergeTwoLists(ListNode l1, ListNode l2) { ListNode dumy = new ListNode(0); Li
