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. 题目要求: 合并两个有序链表 注意: 不能开辟新的结点空间 解题思路: 1.归并排序,创建一个新的头结点,从头到尾分别遍历两个链表,并依次比较其大小关系,每次将头指针指向小的那个. 2.递归思想(对于为改变链表结构的…

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. 分析 数据结构与算法的链表章节的典型实例,将两个有序链表合成一个,保持其有序的性质. AC代码 /** * Definition for singly-linked list. * struct ListNod…

https://leetcode.com/problems/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. 思路: 考察链表操作,没啥说的. AC代码: /** * Definition for singly-lin…

LeetCode第21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. Example: Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4->4 翻译: 合并两个有序链表并返回…

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 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…

这两天一直也没有顾上记录一下自己做过的题目,回头看看,感觉忘的好快,今天做了一个hard,刚开始觉得挺难得,想了两种方法,一种是每次都从k个list中选取最小的一个,为空的直接跳过,再就是每次合并其中的两个list,直到最终合并完成,这就要用到地柜的方法,还有就是划分,感觉递归的思路比较清晰,就拿地柜的写了,使用递归的方法,这道题目就是以21. Merge Two Sorted Lists  为基础的. 阶梯思路就是,每次都从中间切割lists,直到切割完之后左侧或者右侧为,一个或者两个list…

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…

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 类似于归并2个链表,暴力一点的方法就是,每取出一个list就与以前的list归并返回merge后list,知道所有list merge完成. 但是可惜,这样做会TLE.贴下代码先: /** * Definition for singly-linked list. * struct ListNode {…

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. 注意题目要求合并的时候不能新建节点,直接使用原来的节点,比较简单,代码如下: /** * Definition for singly-linked list. * struct ListNode { * int va…

题目 Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 分析 合并k个有序链表. 我们从数据结构的链表章节学会了如何合并两个链表,针对此题,一个简单的解法便是遍历k次,两个链表合并,合并后的结果与下一个链表合并: 此时时间复杂度为O(nk),遗憾,提交会出现TLE. 所以,必须另觅他法,我们知道排序领域中很多方法都可以提高性能比如合并排序性能为O(n…

一天一道LeetCode系列 (一)题目 Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. (二)解题 合并K个已拍好序的链表.剑指上有合并两个已排好序的链表的算法,那么K个数,我们可以采用归并排序的思想,不过合并函数可能需要修改一下,换成合并两个已排好序的链表的方法.代码如下: /** * Definition for singly-linked…

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. struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; ListNode *mergeTwoL…

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 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 题解: 简单的链表遍历,还可用递归做. Solution…

合并k个已合并链表. 思路:先把链表两两合并,直到合并至只有一个链表 /** * 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*>& list…

1.题目描述 2.分析 利用 vector 存储指针,同时合并k个链表. 3.代码 ListNode* mergeKLists(vector<ListNode*>& lists) { vector<ListNode*> p; for (vector<ListNode*>::iterator it = lists.begin(); it != lists.end(); it++) { if (*it != NULL){ ListNode *pn = *it; p.…

1.题目描述 2.问题分析 使用合并两个链表的方法,逐次合并,效率较低.可以考虑同时合并K个链表. 3.代码 ListNode* mergeKLists(vector<ListNode*>& lists) { ListNode *res = NULL; for (vector<ListNode*>::iterator it = lists.begin(); it != lists.end(); it++) { res = mergeTwoLists(res, *it); }…

题目描述:把k个排序的链表组成的列表合并成一个排序的链表 思路: 使用堆排序,遍历列表,把每个列表中链表的头指针的值和头指针本身作为一个元素放在堆中: 第一步中遍历完列表后,此时堆中最多会有n个元素,n是列表的长度: 当堆不为空,取出堆中的最小值,然后把该值的指针指向下一个元素,并入堆: 第3步可以确保堆永远是o(n)大小的: 堆为空返回头结点就可以了 # Definition for singly-linked list. # class ListNode(object): # def __i…

1.题目描述 2.题目分析 题目要求合并有序的两个链表,要求不能额外申请空间. 3.代码 ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { if( l1 == NULL )return l2; if( l2 == NULL )return l1; if( l1 == NULL && l2 == NULL ) return NULL; ListNode* r = ( l1->val <= l2->val )?l1:l…

题目描述(easy) 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. Example: Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4-…

21. 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. Example: Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4->4 代…

1.题目 21. 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. Example: Input: 1->2->4, 1->3->4 Output: 1->1->2->3->4-&g…

题目链接: https://leetcode.com/problems/merge-k-sorted-lists/?tab=Description Problem: 给出k个有序的list, 将其进行合并得到一个有序的list   对于给出的ListNode[] lists ,可以进行两两合并.divide and conquer  将list分为前后两部分,对前半部分再次进行分半操作,对后半部分进行分半操作,然后将其进行合并操作. 合并操作也就是对两个list进行合并 合并操作可以采用递归算法…

# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 21: Merge Two Sorted Listshttps://oj.leetcode.com/problems/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…

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 =…

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…

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 这道混合插入有序链表和我之前那篇混合插入有序数组非常的相…

21. 合并两个有序链表 21. Merge Two Sorted Lists 题目描述 将两个有序链表合并为一个新的有序链表并返回.新链表是通过拼接给定的两个链表的所有节点组成的. LeetCode21. Merge Two Sorted Lists 示例: 输入: 1->2->4, 1->3->4 输出: 1->1->2->3->4->4 Java 实现 ListNode 类 class ListNode { int val; ListNode n…