需要遍历 list2 中的全部 O(|list2|) 个结点 空间复杂度:O(1) 只需要维护常数个额外变量 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, list...
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: 于是乎,新建一个链表,next用两个链表当前位置去比较,...
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,用于存放比较后的结果。然后对传入的两个链表...
的链表连接起来 最后return的是new_list.next,因为new_list是根节点,new_list.next才是头节点! 二、用递归的解法 classListNode:def __init__(self, x): self.val=x self.next=NoneclassSolution:defmergeTwoLists(self, l1, l2):""":type l1: ListNode :type l2: ListNode :rtype: ListNode""" if ...
python 3个list合并为一个二维数组,最近在研究一些算法为蓝桥杯和毕业时的面试做准备,以及增加自己在数据结构方面的理解合并两个有序链表首先是创建一个链表类classListNode():def__init__(self,val,next=None):self.val=valself.next=next1.非递归的写法将两个链表的头部
如果不成立,则将链表 l1 和链表 l2 的下一个节点传入递归调用的 mergeTwoLists 方法,并将返回的结果赋值给链表 l2 的下一个节点,并返回链表 l2。 完整代码 代码语言:javascript 代码运行次数:0 运行 AI代码解释 class Solution(object): def mergeKLists(self, lists): """ :type lists: List[ListNode] :...
defmergeTwoLists(self, list1, list2): if not list1: return list2 if not list2: return list1 result = None while list1 or list2: if not (list1 and list2): while list1: result = ListNode(list1.val,result) list1 = list1.next ...
传入的参数为两个列表,list1准备作为key,list2准备作为value,key和value位置一一对应。 代码语言:javascript 代码运行次数:0 运行 AI代码解释 def list_dic(list1,list2): ''' two lists merge a dict,a list as key,other list as value :param list1:key :param list2:value :return:dict ''' dic ...
This operator checks the equality of elements between two lists. If all elements are the same in the same order, the comparison will return “Equal”. Otherwise, it will return “Not equal”.if my_list1 == my_list2: print("Equal") else: print("Not equal") # Not equal...
public class Test { static class Person { String name; String occupation; } public static void main(String[] args) { String[] nameArray = {"小王", "小张", "小赵"}; String[] occupationArray = {"经商", "学生", "士兵"}; assert nameArray.length == occupationArray.length; List<List...