LeetCode——Add Two Numbers

leetcode上第二题：

题目内容：

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

题目翻译：

给定两个链表，它们代表两个非0的数，数字以逆序存在链表中，链表的每个结点包含一个单一的数。把这两个整数相加并以链表的形式返回。

解题思路：

主要考察链表的操作，两数的各位相加时注意进位。最后相加时注意最高位的值是否为1（即次高位想加时是否产生进位）感觉没有考察具体的算法。

<span style="font-size:14px;">/** * 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 rootNode(0);ListNode *pCurNode = &rootNode;int index = 0;while(l1||l2){int a = (l1?l1->val:0);int b = (l2?l2->val:0);int sum = a+b+index;index = sum/10;sum %= 10;ListNode *newNode = new ListNode(sum);pCurNode->next = newNode;pCurNode = newNode;if(l1)l1 = l1->next;if(l2)l2 = l2->next;}if(index>0){ListNode *newNode = new ListNode(index);pCurNode->next = newNode;}return rootNode.next;    }};</span>