leetcode

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

题目意思很简单，有点类似两个链表归并

个人思路：

1，同时从头结点开始遍历两个链表，将对应的两个结点以及进位相加，和记为temp，进位记为flag，temp % 10是结果链表对应结点的值，temp / 10是进位flag

2，不断循环，直到某个链表遍历完，然后将未遍历完链表的剩余部分与进位相加，处理方式同1，直到该链表也遍历完

3，注意一下最后的进位flag是否为1，为1则表示最后结点的和是有进位的，需要再向高位进1

代码：

#include <stddef.h>

struct ListNode
{
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        if (!l1)
        {
            return l2;
        }

        if (!l2)
        {
            return l1;
        }

        ListNode *l1Current = l1;
        ListNode *l2Current = l2;
        ListNode *result = NULL;
        ListNode *head = NULL;
        int flag = 0; //进位
        int temp = 0;

        while (l1Current && l2Current)
        {
            temp = l1Current->val + l2Current->val + flag;
            if (result)
            {
                result->next = new ListNode(temp % 10);
                result = result->next;
                flag = temp / 10;
            }
            else
            {
                result = new ListNode(temp % 10);
                head = result;
                flag = temp / 10;
            }
            l1Current = l1Current->next;
            l2Current = l2Current->next;
        }

        while (l1Current)
        {
            temp = l1Current->val + flag;
            result->next = new ListNode(temp % 10);
            result = result->next;
            flag = temp / 10;
            l1Current = l1Current->next;
        }

        while (l2Current)
        {
            temp = l2Current->val + flag;
            result->next = new ListNode(temp % 10);
            result = result->next;
            flag = temp / 10;
            l2Current = l2Current->next;
        }

        if (flag) //最后的结点和有进位
        {
            result->next = new ListNode(1);
        }

        return head;
    }
};

网上基本是这个思路