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.
Example
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
解题思路：
1、此题考查用linkedlist模拟加法运算。
2、最直接的想法就是将两个加数恢复成int类型再相加，将结果转换成linkedlist返回。
3、但是int类型允许的整数长度是瓶颈，经过测试发现题目里的测试样例确实有特别长的甚至超过long int类型，那么用意很明显了。
4、只能用linkedlist完全模拟加法运算过程，取数相加，进位，下两位和进位数再相加，处理进位，循环到null...
5、注意：
a. 两个list可能不一样长。
b. 别忘了最后一个非零的进位数。
解题源码：
#include<iostream>
#include<math.h>
using namespace std;
struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
int sum = l1->val + l2->val;
int value = sum % 10, c = sum / 10;
ListNode *res = new ListNode(value);
ListNode *p1 = l1, *p2 = l2, *pr = res;
while(p1->next != NULL || p2->next != NULL)
{
int add_1 = 0, add_2 = 0;
if (p1->next != NULL) {
p1 = p1->next;
add_1 = p1->val;
}
if (p2->next != NULL) {
p2 = p2->next;
add_2 = p2->val;
}
sum = add_1 + add_2 + c;
value = sum % 10, c = sum / 10;
pr->next = new ListNode(value);
pr = pr->next;
}
if (c != 0) {
pr->next = new ListNode(c);
}
return res;
}
};
