## 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 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 ...
1、题目描述 2、题目分析 题目要求合并有序的两个链表,要求不能额外申请空间。 3、代码 1ListNode* mergeTwoLists(ListNode* l1, ListNode*l2) {2if( l1 == NULL )returnl2;3if( l2 == NULL )returnl1;4if( l1 == NULL && l2 == NULL )returnNULL;5ListNode* r = ( l1->val <= l2->val )...
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...
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。
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 ...
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 =...
1. 题目描述 Merge two sorted linked lists and return it as a new sorted list. The new list ...
# @lc app=leetcode id=88 lang=python # # [88] Merge Sorted Array ## @lc code=start class Solution(object): def merge(self, nums1, m, nums2, n): """ :type nums1: List[int] :type m: int :type nums2: List[int] :type n: int ...
leetcode 21. Merge Two Sorted Lists 2019-12-15 01:01 −合并两个已经排好序的链表,注意需要用已有的节点组成新链表。这题与第4题很相似。 合并两个数组如下 ```javascript var newArray = [] function merge(el) { newArray.push(el) } while (true) { ... ...