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

Solution
思路并不難想义起，從頭節(jié)點(diǎn)開(kāi)始纹安，逐個(gè)節(jié)點(diǎn)相加元践，并保存進(jìn)位格二，計(jì)算下一個(gè)節(jié)點(diǎn)時(shí)計(jì)算進(jìn)位。
不過(guò)這一題需要考慮較全面，很容易漏掉處理一些邊界條件，這題提交了5次才通過(guò)狱庇。
對(duì)于一些看似簡(jiǎn)單的題目惊畏，一定要更加小心，不要陰溝翻船密任。
下面代碼中的測(cè)試Case就是教訓(xùn)颜启。

Code
common頭文件的代碼比較簡(jiǎn)單，就不貼上了浪讳。

#include <iostream>                                                                                                                                           

#include "../common/list/list.h"

class Solution
{
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) 
    {   
        if (l1 == NULL && l2 == NULL)
        {
            return new ListNode(0);
        }
        if (l1 == NULL)
        {
            return l2; 
        }
        if (l2 == NULL)
        {
            return l1; 
        }
        ListNode* result = NULL;
        ListNode* pL1 = l1; 
        ListNode* pL2 = l2; 
        ListNode* pResult = NULL;
        int flag = 0;
        while (pL1 != NULL && pL2 != NULL)
        {
            int curNodeVal = (pL1->val + pL2->val + flag) % 10; 
            flag = (pL1->val + pL2->val + flag) / 10; 
            ListNode* curNode = new ListNode(curNodeVal);
            if (result == NULL)
            {
                result = curNode;
                pResult = curNode;
            }
            else
            {
                pResult->next = curNode;
                pResult = curNode;
            }
            pL1 = pL1->next;
            pL2 = pL2->next;
        }
        while (pL1 != NULL)
        {
            int curNodeVal = (pL1->val + flag) % 10; 
            flag = (pL1->val + flag) / 10; 
            ListNode* curNode = new ListNode(curNodeVal);
            pResult->next = curNode;
            pResult = pResult->next;
            pL1 = pL1->next;
        }
        while (pL2 != NULL)
        {
            int curNodeVal = (pL2->val + flag) % 10;
            flag = (pL2->val + flag) / 10;
            ListNode* curNode = new ListNode(curNodeVal);
            pResult->next = curNode;
            pResult = pResult->next;
            pL2 = pL2->next;
        }
        if (flag)
        {
            ListNode* curNode = new ListNode(flag);
            pResult->next = curNode;
        }
        return result;
    }
};

int main()
{
    /* Case: 1
    ListNode* l1 = new ListNode(2,
                                new ListNode(4,
                                             new ListNode(3)));
    ListNode* l2 = new ListNode(5,
                                new ListNode(6,
                                             new ListNode(4)));
    */

    /* Case: 2
    ListNode* l1 = new ListNode(5);
    ListNode* l2 = new ListNode(5);
    */

    /* Case: 3
    ListNode* l1 = new ListNode(9,
                                new ListNode(9));
    ListNode* l2 = new ListNode(1);
    */

    /* Case: 4
    ListNode* l1 = new ListNode(0);
    ListNode* l2 = new ListNode(7,
                                new ListNode(8));
    */

    /* Case 5 */
    ListNode* l1 = new ListNode(1);
    ListNode* l2 = new ListNode(9,
                                new ListNode(9));
    Solution solution;
    ListNode* result = solution.addTwoNumbers(l1, l2);
    PrintList(result);
    return 0;
}