标签：注意 简单的 red 加法 sts blog result ora tput
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
就是简单的高精度加法，刚开始理解错题意WA了好几次。
要注意的是：
判断两个list有没有都算到头；
判断进位是否为0；
考虑有没有多申请节点，最后一个节点不能为0。
以下是我的代码：
/** * 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* re = new ListNode(0); ListNode* result = re; int carry = 0; while (l1 || l2 || carry) { int tmp = (l1?(l1->val):0) + (l2?(l2->val):0) + carry; re->val = tmp % 10; carry = tmp / 10; if (l1) l1 = l1->next; if (l2) l2 = l2->next; if (l1 || l2 || carry) re->next = new ListNode(0), re = re->next; } return result; } };
标签：注意 简单的 red 加法 sts blog result ora tput
原文地址：http://www.cnblogs.com/zmj97/p/7526183.html
