Add the two numbers and return it as a linked list. You may assume the two numbers do not contain any leading zero, except the number 0 itself. 翻译 给定两个非空的链表,代表两个非负整数。这两个整数都是倒叙存储,要求返回一个链表,表示这两个整数的和
最终代码: /*** Definition for singly-linked list.* public class ListNode {* public int val;* public ListNode next;* public ListNode(int x) { val = x; }* }*/publicclassSolution{publicListNodeAddTwoNumbers(ListNodel1,ListNodel2){ListNodehead=newListNode(0);ListNodel3=head;ListNodel4=l3;int...
* Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ publicclassSolution { publicListNode addTwoNumbers(ListNode l1, ListNode l2) { //如果给出就为空,则直接返回另外一个链表 if(l1 ==null)returnl2; if(l...
题目不难,直接上代码。 /*** Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * }*/classSolution {publicListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode result=null, tmp =null;//tmp作为指针使用,进行...
LeetCode 2 Add Two Numbers——用链表模拟加法 点击上方蓝字,和我一起学技术。 今天要讲的是一道经典的算法题,虽然不难,但是很有意思,我们一起来看下题目: You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their ...
You may assume the two numbers do not contain any leading zero, except the number 0 itself. Example: Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8 Explanation: 342 + 465 = 807. 2 词汇学习 non-empty非空non-negative非负reverse相反 ...
Can you solve this real interview question? Add Two Numbers - 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 contains a single digit. Add the two numbers an
for (auto iter = nums.begin(); iter != nums.end();) { current->val = *iter; if (++iter != nums.end()) current->next = new ListNode(0); else current->next = nullptr; // 如果是最后一个数据了,则next指向为NULL current = current->next; ...
1classSolution {2public:3ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {4ListNode *res =newListNode(-1);5ListNode *cur =res;6intcarry =0;7while(l1 ||l2) {8intn1 = l1 ? l1->val :0;9intn2 = l2 ? l2->val :0;10intsum = n1 + n2 +carry;11carry = sum /10;12cur-...
3 输入与输出：/*** Definition for singly-linked list.* struct ListNode {* int val;* ListNode *next;* ListNode(int x) : val(x), next(NULL) {}* };*/class Solution {public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { }};4 解决思路：从表头开始相加，记录每次相加...