【LeetCode】33.Linked List — Add Two Numbers两数相加 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
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 return the sum as a linked list. You may assume the two numbers do not contain any leading ze...
https://leetcode.com/problems/add-two-numbers/ 题目: 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 digit. Add the two numbers and return it as a linked list. Input: (2 -> 4 -> 3...
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 contain a single digit. Add the two numbers and return it as a linked list. ...
The number of nodes in each linked list is in the range[1, 100]. 0 <= Node.val <= 9 It is guaranteed that the list represents a number that does not have leading zeros. 题目大意:2个逆序的链表,要求从低位开始相加,得出结果也逆序输出,返回值是逆序结果链表的头结点。
Leetcode c++ 方法/步骤 1 问题描述：您将获得两个非空链表，表示两个非负整数。 数字以相反的顺序存储，每个节点包含一个数字。 添加两个数字并将其作为链接列表返回。您可以假设这两个数字不包含任何前导零，除了数字0本身。2 问题示例：输入：（2 - > 4 - > 3）+（5 - > 6 - > 4）输出：7 - >...
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 digit. Add the two numbers and return it as a linked list. Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) ...
方法二:遍历两个list,直接按位计算,记录是否进位# Definition for singly-linked list. # class ListNode(object): # def __init__(self, x): # self.val = x # self.next = None class Solution(object): def addTwoNumbers(self, l1, l2): """ :type l1: ListNode :type l2: ListNode :rtype...
* 2)需要额外的 n 大小的空间存储 计算结果结点,空间复杂度为 O(n)。 */ var addTwoNumbers = function(l1, l2) { //定义哨兵结点 let head = new ListNode("head"); let current = head;//临时指针 //存储计算后的链表 let sumNode = head; ...