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...
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...
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 = ...
AI代码解释 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(...
Python练习篇——Leetcode 2. 两数相加(Add Two Numbers) 编程岛 1 人赞同了该文章 注:本文基于64位windows系统(鼠标右键点击桌面“此电脑”图标——属性可查看电脑系统版本)、python3.x(pycharm自动安装的版本, 3.0以上)。 文中代码内容所使用的工具是pycharm-community-2020.1,实践中如有碰到问题,可留言提问...
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 ...
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; ...
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.val = val # self.next = next classSolution(object): defaddTwoNumbers(self,l1,l2): ifnotl1: returnl2 ifnotl2: returnl1 s1,s2='','' whilel1: s1+=str(l1.val) l1=l1.next whilel2: s2+=str(l2.val) l2=l2.next
class Solution { public: ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { // Start typing your C/C++ solution below if(l1 == NULL && l2 == NULL) { return NULL; } ListNode *retList = new ListNode(0); ListNode *tmpList = retList; ...