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. You may assume the two numbers do not contain any leading zero, exc...
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) +...
# 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,方便后续处理 head_pre =...
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
* Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ classSolution { public: ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { intcarryBit = 0; ...
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; 1ListNode* addTwoNumbers(ListNode* l1, ListNode*l2) {2intsum =0;3inti =1;4while(l1 != NULL && l2 !=NULL)5{6sum += i*(l1->val + l2->val);7i *=10;8l...
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
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 ...
* Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { //初始化为0的节点 ...
public class Solution { /* * 方法1 */ public static ListNode addTwoNumbers(ListNode l1,ListNode l2) { //如果都为空 直接返回不为空的一个参数 如果都未空 则返回空 if(l1 == null || l2 == null){ return l1 == null ?(l2 == null ?null:l2):l1; ...