描述: 合并两个有序链表. 解决: ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { if (!l1) return l2; if (!l2) return l1; if (!l1 && !l2) return NULL; ListNode* ret = ); ListNode* now = ret; while (l1 || l2) { if (!l1) { now->next = l2; break; } else if…

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 分析: 将两个有序链表合并为一个新的有序链表并返…

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 list顺序合并就可以了 /** * Definitio…

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,有序链表,递归,迭代,题解,leetcode, 力扣,Python, C++, Java 目录 题目描述 题目大意 解题方法 迭代 Python解法 C++解法 Java解法 递归 日期 题目地址:https://leetcode.com/problems/merge-two-sorted-lists/ 题目描述 Merge two sorted…

这道题是LeetCode里的第21道题. 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. 示例: 输入:1->2->4, 1->3->4 输出:1->1->2->3->4->4 这道题需要考虑的地方挺多的,首先是头节点的处理,还有尾节点的链接问题.对于头节点,我的想法是先对 l1, l2 的值进行比较,然后把 l1 指向头节点值小的,这样保证了 l1 绝对小于等于 l2,也就是说头节点的链接问题…

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 需要排序!!! [2,4][1] 输出1 2 4 而不是2 4 1  递归版!! class Solution { public ListNode mergeTwo…

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 这道题让我们合并k个有序链表,最终合并出来的结果也必须是有序的,之前做过一道 M…

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 题目 合并两个链表 思路 用dummy, 因为需要对头结…

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->4Output: 1->1->2->3->4->4详见:https://leetcode.com/problems…

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 和88. Merge Sorted Array类似,数据…

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…

合并K个有序链表,并且作为一个有序链表的形式返回.分析并描述它的复杂度. 详见:https://leetcode.com/problems/merge-k-sorted-lists/description/ 实现语言:Java /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ cla…

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 解决思路:最简单…

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 思路 这道题最简单的方法就是我们将K个链表中的所有数据都存进数组中,然后对数据进行…

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 要合并K个排好序的链表,我用的方法是用一个优先队列每次存K个元素在队列中,根据优…

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. Input: [ 1->4->5, 1->3->4, 2->6 ] Output: 1->1->2->3->4->4->5->6 题意: 归并k个有序链表. 思路: 用一个最小堆minHeap,将所有链表的头结点放入 新建一个linkedl…

题意:合并两个有序链表 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { if(l1 == NULL) retu…

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…

题目链接 https://leetcode.com/problems/merge-two-sorted-lists/?tab=Description   Problem: 已知两个有序链表(链表中的数值递增排列).将这两个链表进行合并操作,并且满足递增排列 即合并后仍为有序链表.     递归进行合并操作.mergeTwoList(ListNode list1, ListNode list2)      当list1==null时 return list2;      当list2==null时…

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…

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 中文:将两个有序链表合并为一个新的有序链表并返回.新链表…

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. 翻译 合并两个有序的链表 Hints Related Topics: LinkedList 参考 归并排序-维基百科,可以递归也可以迭代,基本的链表操作 代码 Java /** * Definition for…

题目描述: 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. 解题思路: 题目的意思是将两个有序链表合成一个有序链表. 逐个比较加入到新的链表即可. 代码如下: public static ListNode mergeTwoLists(ListNode l1, Li…

题目: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. 简单题,只要对两个链表中的元素进行比较,然后移动即可,只要对链表的增删操作熟悉,几分钟就可以写出来,代码如下: struct ListNode { int val;…

合并链表 Runtime: 4 ms, faster than 100.00% of C++ online submissions for Merge Two Sorted Lists. class Solution { public: ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) { //1 2 4 . 1 3 4 ListNode *res = ); ListNode *cur = res; while (l1 != NULL &&am…

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. 示例: 输入:1->2->4, 1->3->4 输出:1->1->2->3->4->4 class ListNode: def __init__(self, x): self.val = x self.next = None def __repr__(self):# 显示方式 return "{}->{}".format(self.v…

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…

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. 题意:合并两个排好的链表并返回新的链表. 可以使用归并排序,从两链表的表头,取出结点,比较两值,将较小的放在新链表中.如1->3->5->6和2->4->7->8,先将1放入新链表,然后将3…

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. 问题:将两个已排序的列表,合并为一个有序列表. 令 head 为两个列表表头中较小的一个,令 p 为新的已排序的最后一个元素.令 l1, l2 分别为两个列表中未排序部分的首节点.依次将 l1, l2 中的较小值追加…

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. 解法: 新建一个链表,依次比较两个链表的头元素,把较小的移到新链表中,直到有一个为空,再将另一个链表剩余元素移到新链表末尾. 采用循环的方式,代码如下: /** * Definition for singly-lin…