• 描述：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, except the number 0 itself.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

• 分析：求出两个链表中数字的和，其中数组是倒叙存储，要求结果也倒叙输出。

• 思路一：用一个数存储进位，按照链表顺序对数字逐个进行计算。

/**
 * 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) {
        ListNode *result = NULL, *next_node = NULL;
        int adder = 0;
        int l1_val = 0;
        int l2_val = 0;
        while (l1 || l2) {
            int v1 = l1? l1->val: 0;
            int v2 = l2? l2->val: 0;
            int res_val = v1 + v2 + adder;
            int temp = res_val % 10;
            adder = res_val / 10;
            cout << temp;
            ListNode* p = new ListNode(temp); 
            if (!result) result = p;
            if (next_node) next_node -> next = p;
            next_node = p;
            l1 = l1? l1->next: NULL;
            l2 = l2? l2->next: NULL;
        }
        if (adder > 0) {
            ListNode* l = new ListNode(adder);
            next_node -> next = l;
        }
        return result;
    }
};

• 总结：这道题解法有很多种，最直接的就是把数字读出来之后进行相加求解，但是一位一位计算会简单很多。对于结果链表的构造要仔细思考，否则很容易构造错。