Can you solve this real interview question? Add Two Numbers II - You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contains a single digit. Add the two numbers
You are given two linked lists representing two non-negative numbers. The most significant digit comes first and each of their nodes contain a single digit. 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 numbe...
显然此题用到的是stack。 1classSolution {2public:3ListNode* addTwoNumbers(ListNode* l1, ListNode*l2) {4stack<int>s1, s2;5while(l1) {6s1.push(l1->val);7l1 = l1->next;8}9while(l2) {10s2.push(l2->val);11l2 = l2->next;12}13intsum =0;14ListNode* pre =nullptr;15while(!s1....
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。 样例 Input: (2 -> 4 -> 3) +...
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 and retu…
You are given two linked lists representing two non-negative numbers. The most significant digit comes first and each of their nodes contain a single digit. 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 numbe...
LeetCode(2):Add Two Numbers 两数相加 Medium! 题目描述: 给定两个非空链表来表示两个非负整数。位数按照逆序方式存储,它们的每个节点只存储单个数字。将两数相加返回一个新的链表。 你可以假设除了数字 0 之外,这两个数字都不会以零开头。 示例:
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 ...
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...
add_two_numbers( l1: Option<Box<ListNode>>, l2: Option<Box<ListNode>>, )->Option<Box<ListNode>>{ // 声明变量 let mut head = ListNode::new(0); let mut cur = &mut head.next; let (mut x,mut y) = (l1,l2); let mut carry ...