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. Yo
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 ...
def addTwoNumbers(l1: List[Int], l2: List[Int]): List[Int] = { def addWithCarry(lists: (List[Int], List[Int]), carry: Int) = lists match { case (Nil, Nil) => if (carry == 0) Nil else List(carry) case (x :: xtail, Nil) => addHeads(x, 0, carry, (xtail, Nil))...
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: answer: class Solution { public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { ListNode * resultP = new ListNode(0) ; ListNode * head = resultP; ListNode * pre ; while((l1 != NULL) && (l2 != NULL)){ int jinwei...
Next } // 返回结果链表的头结点 return head_pre.Next } 题目链接: Add Two Numbers : leetcode.com/problems/a 两数相加: leetcode-cn.com/problem LeetCode 日更第 55 天,感谢阅读至此的你 欢迎点赞、收藏鼓励支持小满 编辑于 2022-03-12 22:10...
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…
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...
【Leetcode】002— Add Two Numbers Gaoyt__关注赞赏支持【Leetcode】002— Add Two Numbers Gaoyt__关注IP属地: 吉林 0.1522019.06.23 23:57:24字数98阅读149 一、题目描述给出两个非空的链表用来表示两个非负的整数。其中,它们各自的位数是按照 逆序 的方式存储的,并且它们的每个节点只能存储 一位 数字。
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { int sum = l1->val + l2->val; int value = sum % 10, c = sum / 10; ListNode *res = new ListNode(value); ListNode *p1 = l1, *p2 = l2, *pr = res; while(p1->next != NULL || p2->next != NULL) ...
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