2. Add Two Numbers--2016/09/18

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

思路：按照加法顺序依次加链表并保留进位。尤其注意以下两种场景：
1.两链表个数不同，有一个剩的且此时进位为1（递归计算）
2.两链表个数相同，但计算完时进位为1

/** * 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) {        int carry = 0;        ListNode* head = new ListNode(0);        ListNode* curNode = head;        ListNode* carryNode = new ListNode(1);        int num1 = 0;        int num2 = 0;        int sum = 0;        //计算加法        while(l1 && l2)        {            num1 = l1->val;            num2 = l2->val;            sum = carry + num1 + num2;            carry = sum/10;            curNode->next = new ListNode(sum%10);            curNode = curNode->next;            l1 = l1->next;            l2 = l2->next;        }        //其中某一个list未走完        if(l1)        {            //注意进位没用完            if(carry == 1)            {                   curNode->next = addTwoNumbers(carryNode,l1);            }            else            {                curNode->next = l1;            }        }        else        {            if(l2)            {                //注意进位没用完                if(carry == 1)                {                       curNode->next = addTwoNumbers(carryNode,l2);                }                else                {                    curNode->next = l2;                }            }            else            {                //注意两个list都走完了，但是进位为1                if(l1 == NULL && l2 == NULL && carry == 1)                {                    curNode->next = new ListNode(1);                }            }        }        curNode = head->next;        free(carryNode);        free(head);        return curNode;    }};