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
这道并不是什么难题，算法很简单，链表的数据类型也不难。就是建立一个新链表，然后把输入的两个链表从头往后撸，每两个相加，添加一个新节点到新链表后面，就是要处理下进位问题。还有就是最高位的进位问题要最后特殊处理一下。代码如下：
/** * 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) { ListNode *res = new ListNode(0); ListNode *cur = res; int carry = 0; while (l1 || l2) { int n1 = l1 ? l1->val : 0; int n2 = l2 ? l2->val : 0; int sum = n1 + n2 + carry; if (sum > 9) { carry = 1; sum -= 10; } else carry = 0; cur->next = new ListNode(sum); cur = cur->next; if(l1) l1 = l1->next; if(l2) l2 = l2->next; } if (carry == 1) cur->next = new ListNode(1); return res->next; } };
[LeetCode] Add Two Numbers 两个数字相加
原文：http://www.cnblogs.com/grandyang/p/4129891.html