## 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...
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...
next = list2; } // 返回合并后的链表的头结点 head_pre.next } } 题目链接: 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...
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 ``` ###拿到手的第一想法 实话说,这道题看到是链表类型的时候,我自己...
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) {...
leetcode 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...
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 ...
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。
LeetCode 23. Merge k Sorted Lists 经典题 解法一:优先队列 {{{ class CMP{ public: bool operator () (const ListNode* a, const ListNode * b) const { return a->val > b->val; } }; class Solution { public: ListNode* mergeKLists(vector<ListNode*>& lists) { ListNode *head = nullptr;...