题目链接: https://oj.leetcode.com/problems/merge-k-sorted-lists/ 问题: Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 解题思路: 方法一:暴力破解 思路: 将列表一个一个地合并(例如:lists = [l1, l2, l3, l4],首先合并l1和l2,然后将合并后...
Merge two sorted linked lists and return it as a sorted list. The list should be made by splicing together the nodes of the first two lists. Example 1:Input:l1=[1,2,4],l2=[1,3,4]Output:[1,1,2,3,4,4] Example 2:Input:l1=[],l2=[]Output:[] Example 3:Input:l1=[],l2=[0...
改进mergeTwoLists方法,以在开始时检查空链表。 class ListNode: def __init__(self, x): self.val = x self.next = None # 改进后的将给出的数组转换为链表的函数 def linkedlist(list): if not list: # 检查列表是否为空 return None # 空列表返回None head = ListNode(list[0]) cur = head for...
需要遍历 list2 中的全部 O(|list2|) 个结点 空间复杂度:O(1) 只需要维护常数个额外变量 代码(Python3) # Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def mergeTwoLists(self, list...
Merge two sorted linked lists and return it as a sorted list. The list should be made by splicing together the nodes of the first two lists. Example 1: Input: l1 = [1,2,4], l2 = [1,3,4] Output: [1,1,2,3,4,4] Example 2: ...
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 提议很简单,就是归并排序。 首先想到的即使逐个归并得到最终的结果,但是会超时,这是因为这种会造成数组的size大小不一样,导致归并排序的时间变长; ...
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. 题目大意:合并两个有序的链表 思路:通过比较两个链表的节点大小,采用尾插法建立链表。 代码如下: ...
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:合并两个排好序的链列,并将其作为新链表返回。新链表应通过将前两个列表的节点拼接在一起。
用一个大小为K的最小堆(用优先队列+自定义降序实现)(优先队列就是大顶堆,队头元素最大,自定义为降序后,就变成小顶堆,队头元素最小),先把K个链表的头结点放入堆中,每次取堆顶元素,然后将堆顶元素所在链表的下一个结点加入堆中。 代码语言:javascript ...
一、问题描述 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->4->4->5->6 二、解决思路 思路一:直接通过循环,每次求出链表数组中最小节点直到结束 ...