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 addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: # 哨兵结点,方便后续处...
最终成绩是: 第二种方法是直接在链表上面进行操作,两个数直接相加,后面再进行处理 # Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def addTwoNumbers(self, l1: ListNode, l2: ListNode) -...
class Solution(object): def addTwoNumbers(self, l1, l2): """ :type l1: ListNode :type l2: ListNode :rtype: ListNode """<br>#把链表值放进列表中。方便之后迭代 l1a = [] l2a = [] result = [] l1a.append(l1.val) l2a.append(l2.val) loop = ListNode(0) loop1 = l1 loop2 = ...
You are given two linked lists representing two non-negative numbers.The digits are stored in reverse order and each of their nodes contain a single digitAdd the two numbers and return it as a linked listInput: (2->4->3) + (5->6->4)Output: 7->0->8代码如下,递归调用add函数即可:1...
classSolution{public:ListNode*addTwoNumbers(ListNode*l1,ListNode*l2){ListNode*ret=newListNode();ListNode*pnt=ret;bool carry=false;while(l1!=nullptr||l2!=nullptr||carry){int cur=0;if(l1!=nullptr){cur+=l1->val;l1=l1->next;}if(l2!=nullptr){cur+=l2->val;l2=l2->next;}if(carry){cur...
class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { //初始化为0的节点 ListNode init = new ListNode(0); //当前输出节点位置,初始化为initListNode ListNode current = init; //进位值 int carry = 0; while (l1 != null || l2 != null) { ...
class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode dummy = new ListNode(0); ListNode cur = dummy; int carry = 0; while (l1 != null || l2 != null || carry != 0) { int i1 = l1 != null ? l1.val : 0; ...
思路展示图片、答案思路来源: https://leetcode-cn.com/problems/add-two-numbers/solution/liang-shu-xiang-jia-by-leetcode/ 我们将跟踪逐位计算过程,逐位相加满 10 取余数作为结果,并进位 1 给下一节点相加,逐位将结果连为链表。 代码语言:javascript 代码运行次数:0 运行 AI代码解释 class Solution: def ...
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...
classSolution{ public: vector<vector<int>>threeSum(vector<int>&nums) { } }; 已存储 行1,列 1 运行和提交代码需要登录 Case 1Case 2Case 3 nums = [-1,0,1,2,-1,-4] 9 1 2 3 › [-1,0,1,2,-1,-4] [0,1,1] [0,0,0] ...