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...
以下是Python代码,并有测试过程。 #coding:utf-8#Definition for singly-linked list.classListNode(object):def__init__(self, x): self.val=x self.next=NoneclassSolution(object):defmergeTwoLists(self, l1, l2):""":type l1: ListNode :type l2: ListNode :rtype: ListNode"""ifnotl1andnotl2:ret...
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...
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...
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
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) { ...
public ListNode mergeTwoLists(ListNode l1, ListNode l2) { if (l1 == null) return l2; if (l2 == null) return l1; if (l1.val < l2.val) l1.next = mergeTwoLists(l1.next, l2); else l2.next = mergeTwoLists(l1, l2.next); ...
as you can see its the same procedure. Put the two lists together, sort and output. Hope this helps, R. You problem is a pretty basic Python problem: "Iterate trough two lists in parallel. Add the items of both list sequentially to the list of results." ...
首先,merge()函数调用前必须保L1,L2是有序的,然后才能调用merge()函数。 所以应该做两处更改: 1.在worker类的公有函数中添加以下 定义排序准则的函数。 bool operator<(const worker & kk) { return this->age < kk.getAge(); } 2.在调用L1.merge(L2)的前面添加以下两行。 L1.sort(); L2.sort()...
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...