21. Merge Two Sorted Lists —— Python 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. 没事来做做题,该题目是说两个排序好的链表组合起来,依然是排序好的,即链表的值从小到大。 代码: 于是...
21. Merge Two Sorted Lists (python) 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 利用链表的思想,先创建个空链表p,用于...
Python代码: AI检测代码解析 # Definition for singly-linked list. class ListNode(object): def __init__(self, x): self.val = x self.next = None class Solution(object): def mergeTwoLists(self, l1, l2): """ :type l1: ListNode
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. 合并两个有序链表 /** * Definition for singly-linked list. * struct ListNode { * int val; ...
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->4Output:1->1->2->3->4->4 ...
获取元素数大于 1 的列表的中点。使用Python语言时,//执行除法,不带余数。它将除法结果四舍五入到最接近的整数。这也被称为楼层划分。 使用中点作为参考点,将列表拆分为两半。这是分而治之算法范例的分而治之的一面。 Recursion is leveraged at this step to facilitate the division of lists into halved co...
用一个大小为K的最小堆(用优先队列+自定义降序实现)(优先队列就是大顶堆,队头元素最大,自定义为降序后,就变成小顶堆,队头元素最小),先把K个链表的头结点放入堆中,每次取堆顶元素,然后将堆顶元素所在链表的下一个结点加入堆中。 代码语言:javascript ...
Merge all the linked-lists into one sorted linked-list and return it. Input and output examples Example 1:Input: lists = [[1,4,5],[1,3,4],[2,6]]Output: [1,1,2,3,4,4,5,6]Explanation: The linked-lists are:[ 1->4->5, 1->3->4, 2->6]merging them into one sorted ...
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...
改进mergeTwoLists方法,以在开始时检查空链表。 class ListNode: def __init__(self, x): self.val = x self.next = None # 改进后的将给出的数组转换为链表的函数 def linkedlist(list): if not list: # 检查列表是否为空 return None # 空列表返回None head = ListNode(list[0]) cur = head for...