当list1 和 list2 均还有结点时,取它们中较小的头结点放入结果链表中,然后不断循环。 最后当其中一个链表为空时,将另一个链表剩余的部分全部插入结果链表尾部即可。 进阶: LeetCode 23: 合并 k 个有序链表 LeetCode 148: 对无序链表排序 时间复杂度:O(|list1| + |list2|) 需要遍历 list1 中的全部 O
1 Linked List 和 Python 基础操作 1.1 链表原理 1.2 链表的操作:增删查改 1.3 链表 in Python 2 LeetCode 21 合并两个有序链表 2.1 读题 2.2 完整的代码实现(包括了前面的结点定义、数组和链表转换函数) 时间复杂度分析 空间复杂度分析 考虑空列表的代码改进 ...
packageleetcode;importjava.util.ArrayList;importjava.util.List;/** * 合并两个已排序的链接列表并将其作为新列表返回。 新列表应该通过拼接前两个列表的节点来完成。 *例: * 输入:1-> 2-> 4, 1-> 3-> 4 * 输出:1-> 1-> 2-> 3-> 4-> 4 *@author小川94 *@date2018-10-21 */publicclass...
Input:1->2->4, 1->3->4Output:1->1->2->3->4->4 因为没有空间要求,所以想到ListNode*head = new ListNode(INT_MIN);重新定义一个链表,分别比较两个有序链表的大小然后将所在结点一次加入到定义的新链表中。 最后注意释放头结点空间。 /** * Definition for singly-linked list. * struct ListNode...
LeetCode-Java-21. Merge Two Sorted Lists 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
https://leetcode.com/problems/merge-two-sorted-lists/ 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: easy 。 算法: 1. public ListNode mergeTwoLists(ListNode l1, ListNode l2) {...
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:合并两个原本有序的列表 c++代码: /// main.cpp// LeetCode/// Created by linSir on 2017/9/5.// Copyright © 2017年 58qifu. All ...
1. 题目描述 Merge two sorted linked lists and return it as a new sorted list. The new list ...
/** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) { //首先判断有没有空链表的情况。。。 if(l1 && !l2) return l1; if(!l1 && l2) return l2; if(!l...
Can you solve this real interview question? Merge Two Sorted Lists - You are given the heads of two sorted linked lists list1 and list2. Merge the two lists into one sorted list. The list should be made by splicing together the nodes of the first two li