代码实现: 1classSolution {2public:3ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {4ListNode preHead(0), *p = &preHead;5intcarry =0;6while(l1 || l2 ||carry) {7intsum = (l1 ? l1->val :0) + (l2 ? l2->val :0) +carry;8carry = sum /10;9p->next =newListNode(sum %10);10p = p->next...
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) +...
public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode result = new ListNode(0); ListNode tem = l1; ListNode tem2 = l2; ListNode current = result; boolean flag = false; while (tem != null || tem2 != null) { // 计算节点值之和,需要判空 int sum = (tem != null ?
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...
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode: link1 = l1 link2 = l2 while(link1!=None and link2 !=None): a = link1.val link2.val += a link1.val = link2.val link1 = link1.next link2 = link2.next ...
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 non-empty You may assume the two numbers do not contain any leading zero, except the number 0 itself. Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: answer: class Solution { public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {...
* 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; ...
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...
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) ...