Python 中也没有链表这个概念和结构,是要自己定义吗?尤其是看到答题区默认的格式: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 # Definitionforsingly-linked list.#classListNode:# def__init__(self,x):# self.val=x # self.next=NoneclassSolution:defaddTw
题目数据保证列表表示的数字不含前导零 题解一(python): 1 # Definition for singly-linked list. 2 # class ListNode: 3 # def __init__(self, val=0, next=None): 4 # self.val = val 5 # self.next = next 6 class Solution: 7 def addTwoNumbers(self, l1: ListNode, l2: ListNode) ->...
python代码段 #Definition for singly-linked list.#class ListNode(object):#def __init__(self, x):#self.val = x#self.next = NoneclassSolution(object):defaddTwoNumbers(self, l1, l2):""":type l1: ListNode :type l2: ListNode :rtype: ListNode"""tag=0ifl1.val+l2.val>9: l3= ListNode...
class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { Stack<Integer> st1 = new Stack(); Stack<Integer> st2 = new Stack(); while (l1 != null) { st1.push(l1.val); l1 = l1.next; } while (l2 != null) { st2.push(l2.val); l2 = l2.next; } int carry...
class Solution: def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode: link1 = l1 link2 = l2 while(link1!=None and link2 !=None): a = link1.val link2.val += a link1.val = link2.val link1 = link1.next
# class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def addTwoNumbers(self, l1, l2): """ :type l1: ListNode :type l2: ListNode :rtype: ListNode """ ret = None tmp = None total = 0 while True: a = l1.val if l1 else 0 b...
nullptr : l2->next; } return ret; }};Python Code:# Definition for singly-linked list.# class ListNode:# def __init__(self, x):# self.val = x# self.next = Noneclass Solution: def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode: def listToSt...
# self.next=nextclassSolution:defaddTwoNumbers(self,l1:ListNode,l2:ListNode)->ListNode:# 定义两个栈用于存储链表中的节点值 stack1,stack2=[],[]# 最后的结果链表 ans=None # 进位 carry=0whilel1:stack1.append(l1.val)l1=l1.nextwhilel2:stack2.append(l2.val)l2=l2.nextwhilestack1 or stack2...
You may assume that each input would have exactly one solution and you may not use the same element twice. 给定一个已经按升序排序的整数数组,找到两个数字,使它们相加到一个特定的目标数。 函数twoSum应该返回两个数字的索引,使它们相加到目标,其中index1必须小于index2。 请注意,您返回的答案(index1和...
LeetCode solutions in C++ 11 and Python3. NO.TitleSolutionNoteDifficultyTag 0 Two Sum C++ Python Note Easy Mapping 1 Add Two Numbers C++ Python Note Medium LinkedList 2 Longest Substring Without Repeating Characters C++ Python Note Medium Mapping 3 Median of Two Sorted Arrays C++ Python Note Har...