代码(Python3) # Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]: # 使用一个哨兵...
## 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...
21: Merge Two Sorted Lists https://oj.leetcode.com/problems/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. ===Comments by Dabay=== 基本链表的操作。先做一个头节点...
https://leetcode.com/problems/merge-two-sorted-lists/ 题意分析: 题目给出两个排好序的链表,将这两个链表整合成一个新的有序的链表。 题目思路: 这道题目很简单,首先构造一个新的链表,比较两个链表的指针指向的节点的值大小,将值较少的节点放到新链表中,并将指针位置往后移动一位,直到某个链表为空之后,...
21. 合并两个有序链表 Merge Two Sorted Lists【LeetCode 力扣官方题解】 1.1万 2 10:55 App 第15课 教媳妇编程: 合并两个有序列表/链表 387 1 2:32 App leetcode21 Merge Two Sorted Lists 合并有序链表 python 讲解 656 -- 4:23 App Leetcode 0021 合并两个有序链表【递归解法】 6670 16 9...
合并两个有序链表 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) { ...
leetcode-21-Merge Two Sorted Lists,Comparethecurrentvalueofeachlist,theassignthesmallernodetomergedlist,thenreturnit.Itisstraightforward,justneedtoknowthati
python 实现 代码语言:javascript 代码运行次数:0 运行 AI代码解释 classSolution:defmergeTwoLists(self,l1,l2):ifl1 is None:returnl2 elif l2 is None:returnl1 elif l1.val<l2.val:l1.next=self.mergeTwoLists(l1.next,l2)returnl1else:l2.next=self.mergeTwoLists(l1,l2.next)returnl2 ...
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. 依次拼接 复杂度 时间O(N) 空间 O(1) 思路 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。