2-Add Two Numbers

难度：medium
类别：linked list

1.题目描述

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

2.算法分析

注意对头结点首先进行处理，并且在最后的时候记得将head->next设置为NULL即可，详细思路可以见前面写过的博客
http://blog.csdn.net/lichen_yun/article/details/78224990
本题比前面写过的(⊙o⊙)容易，直接在前面那篇博客中将反转部分去掉即可。

3.代码实现

/** * 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) {        stack<ListNode*> mystack;        //  get reversed list l1        ListNode* p = l1;        //  sum correspond num, notice the carry        int carry = 0;        ListNode* head = NULL;        if (l1 != NULL && l2 != NULL) {            carry = (l1->val + l2->val) / 10;            head = new ListNode((l1->val + l2->val)%10);            l1 = l1->next;            l2 = l2->next;        }        else if (l1 != NULL && l2 == NULL) {            head = new ListNode(l1->val);            l1 = l1->next;        }        else if (l1 == NULL && l2 != NULL) {            head = new ListNode(l2->val);            l2 = l2->next;        }        ListNode* result = head;        int sum = 0;        while (l1 != NULL || l2 != NULL) {            if (l1 != NULL && l2 != NULL) {                if (carry == 1) sum = l1->val + l2->val + 1;                else sum = l1->val + l2->val;                carry = sum / 10;                head->next = new ListNode(sum % 10);                l1 = l1->next;                l2 = l2->next;            }            else if (l1 != NULL && l2 == NULL) {                head->next = new ListNode((l1->val + carry)%10);                carry = (l1->val + carry) / 10;                l1 = l1->next;            }            else if (l1 == NULL && l2 != NULL) {                head->next = new ListNode((l2->val + carry)%10);                carry = (l2->val + carry) / 10;                l2 = l2->next;            }            head = head->next;        }        //  if the final carry is one        if (carry == 1) {            head->next = new ListNode(1);            head = head->next;        }        head->next = NULL;        return result;    }};