问题描述：
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
用一个链表表示一个非负数，链表尾的数为最高位，输入两个链表，求相加后的数对应的链表。
解题思路：
这道题很简单，需要注意的是输入的两个链表长度可能不同，
逐位相加。
注意结果的链表头和链表尾（最高位可能有进位）的处理就可以了。
参考代码：
非递归版本： /** * 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 s = l1->val + l2->val; int carry = s / 10; ListNode *head = new ListNode(s % 10); ListNode *r = head; l1 = l1->next; l2 = l2->next; while(l1 || l2 || carry){ int w = carry; if (l1){ w += l1->val; l1 = l1->next; } if (l2){ w += l2->val; l2 = l2->next; } r->next = new ListNode(w % 10); r = r->next; carry = w / 10; } return head; } }; 递归版本： /** * 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) { return add(l1,l2,0); } ListNode* add(ListNode* l1, ListNode* l2, int carry){ if (!l1 && !l2 && !carry)return 0; int s = 0; if (l1)s += l1 -> val; if (l2)s += l2 -> val; s += carry; ListNode *node = new ListNode(s % 10); if (l1)l1 = l1->next; if (l2)l2 = l2->next; node->next = add(l1,l2,s / 10); return node; } }; 