# -*- 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…

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…

21. Merge Two Sorted Lists[easy] 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. 解法一: /** * Definition for singly-linked list. * struct ListNode { * int val…

21.Merge Two Sorted Lists 初始化一个指针作为开头,然后返回这个指针的next class Solution { public: ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { ListNode* dummy = ); ListNode* p = dummy; while(l1 && l2){ if(l1->val <= l2->val){ p->next = l1; p = p-&…

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…

一.题目说明 这个题目是21. Merge Two Sorted Lists,归并2个已排序的列表.难度是Easy! 二.我的解答 既然是简单的题目,应该一次搞定.确实1次就搞定了,但是性能太差: Runtime: 20 ms, faster than 8.74% of C++ online submissions for Merge Two Sorted Lists. Memory Usage: 9.4 MB, less than 5.74% of C++ online submissions…

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…

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…

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…

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…

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-…

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…

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. 没事来做做题,该题目是说两个排序好的链表组合起来,依然是排序好的,即链表的值从小到大. 代码: 于是乎,新建一个链表,next用两个链表当前位置去比较,谁的小就放谁.当一个链表放完之后,说明另外一个链表剩下的…

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. public ListNode mergeTwoLists(ListNode l1, ListNode l2) { ListNode p1 = l1; ListNode p2 = l2; ListNode fakeH…

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.首先判断是否有空链表,如果有,则直接返回另一个链表,如果没有,则开始比较两个链表的当前节点,返回较小的元素作为前驱,并且指针向后移动一位,再进行比较,如此循环,知道一个链表的next指向NULL,将另一个链表的…

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. 代码如下: /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListN…

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. /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(in…

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. 思路:比较每个列表的第一个元素. 合并小的添加到列表中. 最后,当其中一个是空的,只需将它附加到合并后的列表,因为它已经排序. /** * Definition for singly-linked list.…

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…

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…

一天一道LeetCode系列 (一)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. (二)解题 这题是剑指offer上的老题了,剑指上面用的是递归,我写了个非递归的版本. /** * Definition for singly-linked list. *…

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. 递归实现: /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(i…

题目内容: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. 题目分析:本题是要合并两个已经有序的单链表,思路很简单,有两种方法:非递归和递归. 题目代码: (1)非递归: 为方便操作,定义一个辅助的头节点,然后比较原来两个链表的头节点,将小的那一个加入到合并链表,最…

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…

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…

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:合并两个有序序列,这个归并排序中的一个关键步骤.这里是要合并两个有序的单链表.由于链表的特殊性,在合并时只需要常量的空间复杂度. 编码: public ListNode mergeTwoLists(Li…

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 # Definition for singly-link…

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…

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. /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x)…

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. 解题分析: 再基础不过的题了,直接看代码吧^-^ 具体代码: /** * Definition for singly-linked list. * public class ListNode { * in…