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 2.3 解法二:递归 时间复杂度分析 空间
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 因为没有空间要求,所以想到ListNode*head = new ListNode(INT_MIN);重新定义一...
【LeetCode算法-21】Merge Two Sorted Lists 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...
list1 和 list2 均为非降序链表 样例 思路:模拟 由于两个链表已经是升序的,所以可以按题意直接模拟处理。 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...
LeetCode-cn 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] ...
LeetCode-Merge Two Sorted Lists Description: 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...
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. 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...
1. 题目描述 Merge two sorted linked lists and return it as a new sorted list. The new 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. 时间O(N) 空间 O(1) 该题就是简单的把两个链表的节点拼接起来,我们可以用一个Dummy头,将比较过后的节点接在这个Dummy头之后。最后如果有没比较完的,...
LeetCode 21. Merge Two Sorted Lists 合并两个有序链表 1. problem description 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. 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...