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 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->4Output:1->1->2->3->4->4 因为没有空间要求,所以想到ListNode*head = new ListNode(INT_MIN);重新定义一...
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,新链表必须由前两个链表拼接...
# 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]: # 使用一个哨兵结点 head_pre ...
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 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 1. public ListNode mergeTwoLists(ListNode l1, ListNode l2) { ...
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. Seen this question in a real interview before? Yes 简单的归并排序,不说了,直接上代码:
Can you solve this real interview question? Merge Two Sorted Lists - You are given the heads of two sorted linked lists list1 and list2. Merge the two lists into one sorted list. The list should be made by splicing together the nodes of the first two li
作者:LeetCode-Solution链接:https://leetcode-cn.com/problems/merge-two-sorted-lists/solution/he-bing-liang-ge-you-xu-lian-biao-by-leetcode-solu/ 复杂度分析 M、N 是两条链表 l1、l2 的长度 时间复杂度:时间复杂度:O(n + m)O(n+m),其中 nn 和 mm 分别为两个链表的长度。因为每次调用递归都...
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. Input: 1->3->5->null 2->4->6->7->null Output:1->2->3->4->5->6->7->null Day 1944 答案揭晓 DS Interview Question & Answer Expla...