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一个新的链表,新链表必须由...
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 有序链表的合并以前在数据结构课上就上过了,但是现在写起来不是很顺手,可能是因为太久没接触了,当然,整体思路还是很清晰的: ) 对于输入的两个...
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: AI检测代码解析 ...
leetcode -- Merge Two Sorted Lists -- 重点 https://leetcode.com/problems/merge-two-sorted-lists/ 我的思路就是找到第一个j > i的,然后用pre_j串起来。然后相应地对i再做一次。这种思路不太好。 note在指针前进的时候,判断越界的语句要放在if 或者while的前半部分。
文章作者:Tyan 博客:noahsnail.com|CSDN|简书 1. Description Merge Two Sorted Lists 2. Solution /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} ...
1. 题目描述 Merge two sorted linked lists and return it as a new sorted list. The new list ...
1、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. 合并2个有序链表 2、代码实现 ... 1、题目 Merge two sorted linked lists and return it as a new list. The new list should be made...
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. 题目大意:合并两个有序的链表 思路:通过比较两个链表的节点大小,采用尾插法建立链表。
Merge Two Sorted Lists : leetcode.com/problems/m 合并两个有序链表: leetcode-cn.com/problem LeetCode 日更第 52 天,感谢阅读至此的你 欢迎点赞、收藏鼓励支持小满 编辑于 2022-03-09 07:18 力扣(LeetCode) Python Rust(编程语言) 赞同添加评论 分享喜欢收藏申请转载 ...
## 解法二:递归 Recursion class Solution: def mergeTwoLists(self, head1, head2): ## head1 和 head2 头结点指代两个链表 ## 递归的 base,结束条件 if head1 is None: return head2 elif head2 is None: return head1 elif head1.val < head2.val: ## 把小的值 head1 提取出来,拼接上后面...