输入:1->2->4, 1->3->4 输出:1->1->2->3->4->4 解题思路 1.声明一个新链表l;2.判空l1和l2,有空就返回另一个链表;3.对比l1和l2的第一个节点数值,把数值小的加入新链表l;4.重复执行3,直到有一个链表为空,把非空链表加入l5.返回。 代码 class Solution { public: ListNode*mergeTwoLists(...
l1.next=mergeTwoLists(l2,l1.next);returnl1; }else{ l2.next=mergeTwoLists(l1, l2.next);returnl2; } }
next return list class Solution: def mergeTwoLists(self, head1, head2): if not head1: # 如果head1为空,直接返回head2 return head2 if not head2: # 如果head2为空,直接返回head1 return head1 pre = ListNode(0) # 使用哑结点简化操作 head = pre while head1 and head2: if head1.val ...
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 ...
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。 my code: class Solution(object): ...
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 ...
一、题目描述 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的两个链表的所有节点组成的。示例: 二、代码实现
class ListNode: def __init__(self, x): self.val = x self.next = None class Solution: def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode: head = ListNode(0) first = head while l1 and l2: if l1.val > l2.val: head.next = l2 l2 = l2.next else: head.next =...
每日算法——leetcode系列问题 Merge Two Sorted Lists Difficulty: EasyMerge 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; * ...
Can you solve this real interview question? Merge Two Binary Trees - You are given two binary trees root1 and root2. Imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not. You need