}publicListNode AddTwoNumbers(ListNode l1, ListNode l2) { ListNode c1=l1; ListNode c2=l2; ListNode head=newListNode(0); ListNode p=head;intsum =0;while(c1!=null||c2!=null) { sum/=10;if(c1!=null) { sum+=c1.val; c1=c1.next; }if(c2!=null) { sum+=c2.val; c2=c2.next; }...
ListNode* addTwoNumbers(ListNode* l1, ListNode*l2) {intcur=0;intc=0; ListNode* root=newListNode(0); ListNode* res=root;while(l1 &&l2) { res->next=newListNode(0); res=res->next; cur=c+l1->val+l2->val; c=cur/10; cur=cur%10; res->val=cur; l1=l1->next; l2=l2->next; }...
创建链表的方式:head->next = new ListNode((l1_val + l2_val + c) % 10); 通过的||+循环内部if的方式,将长链表超出部分的处理也写在一个循环里面 class Solution { public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { int c = 0; ListNode* head = new ListNode(-1); ListNode* r...
刷题之路第二题--Add Two Numbers 陈zq关注IP属地: 四川 2019.04.07 09:20:52字数612阅读223 问题简介:输入两个数字链表,输出求和后的链表(链表由数字位数倒序组成) 问题详解: 给定两个非空链表,表示两个非负整数. 数字以相反的顺序存储,每个节点包含一位数字.对两个整数作求和运算,将结果倒序作为链表输出. ...
next = None class Solution: def addTwoNumbers(self, l1, l2): return self.addTwoNumbersRecursive(l1, l2, 0) # return self.addTwoNumbersIterative(l1, l2) def addTwoNumbersRecursive(self, l1, l2, c): val = l1.val + l2.val + c c = val // 10 ret = Node(val % 10) if l1....
=v1.end()){t=(*it1+c)%10;c=(*it1+c)/10;res.push_back(t);++it1;}if(c!=0)res.push_back(c);reverse(res.begin(),res.end());while(res[0]==0&&res.size()!=1)res.erase(res.begin());returnres;}classSolution{public:ListNode*addTwoNumbers(ListNode*l1,ListNode*l2){vector<...
Program to add two numbers using function in C++ #include <iostream>usingnamespacestd;//function declarationintaddition(inta,intb);intmain() {intnum1;//to store first numberintnum2;//to store second numberintadd;//to store addition//read numberscout<<"Enter first number: "; cin>>num1;...
2. Add Two Numbers Medium You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list. ...
Add Two Numbers 方法一: 考虑到有进位的问题,首先想到的思路是: 先分位求总和得到 totalsum,然后再将totalsum按位拆分转成链表; 代码语言:javascript 复制 1ListNode*addTwoNumbers(ListNode*l1,ListNode*l2){2int sum=0;3int i=1;4while(l1!=NULL&&l2!=NULL)5{6sum+=i*(l1->val+l2->val);7i*=10...
class Solution(object): def addTwoNumbers(self, l1, l2): head = ListNode(0) ptr = head carry = 0 while True: if l1 != None: carry += l1.val l1 = l1.next if l2 != None: carry += l2.val l2 = l2.next ptr.val = carry % 10 carry /= 10 # 运算未结束新建一个节点用于储...