LeetCode 2 Add Two Numbers

最新推荐文章于 2022-02-22 21:07:37 发布

原创最新推荐文章于 2022-02-22 21:07:37 发布 · 350 阅读

0 ·

CC 4.0 BY-SA版权

【Good Job】同时被 3 个专栏收录

57 篇文章

订阅专栏

[数据结构]

52 篇文章

订阅专栏

数据结构：链表

15 篇文章

订阅专栏

本文提供了一个LeetCode经典题目“两数相加”的解决方案。该题要求将两个以链表形式表示的非负整数相加，并返回结果链表。文中详细介绍了如何通过迭代方式处理链表节点，实现数值相加并处理进位。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

LeetCode

2 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) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

解答

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *addTwoNumbers(ListNode *list1, ListNode *list2)
    {
        if (list1 == NULL)
            return list2;
        if (list2 == NULL)
            return list1;

        int sum = 0, carry = 0;
        ListNode *head = new ListNode(0);

        sum = (list1->val) + (list2->val);
        carry = sum / 10;
        sum = sum % 10;

        head->val = sum;
        head->next = NULL;

        ListNode *headp = head;
        ListNode *list1p = list1->next;
        ListNode *list2p = list2->next;

        while (list1p != NULL || list2p != NULL)
        {
            int val1 = 0, val2 = 0;
            if (list1p != NULL)
            {
                val1 = list1p->val;
                list1p = list1p->next;
            }
            if (list2p != NULL)
            {
                val2 = list2p->val;
                list2p = list2p->next;
            }

            sum = val1 + val2 + carry;
            carry = sum / 10;
            sum = sum % 10;

            headp->next = new ListNode(0);
            headp->next->val = sum;
            headp->next->next = NULL;
            headp = headp->next;
        }

        if (carry > 0)
        {
            headp->next = new ListNode(0);
            headp->next->val = carry;
            headp->next->next = NULL;
        }

        return head;
    }
};