[Leetcode] 2. Add Two Numbers 解题报告

题目：

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
思路：

简单的链表相加，关键问题是如何把代码写的简洁明了。在有关链表的处理中，有时在头部添加一个额外的头结点，可以额外减少很多判断语句，例如下面代码中的head。此外，将while的循环条件置位(l1 || l2)，也可以大大减少很多条件判断语句。

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) {
if(!l1) return l2;
if(!l2) return l1;
ListNode head(0), *p = &head;
int carry = 0;
while(l1 || l2)
{
int sum = 0, val;
if(l1) sum += l1->val, l1 = l1->next;
if(l2) sum += l2->val, l2 = l2->next;
val = (sum+carry)%10, carry = (sum+carry)/10;
p->next = new ListNode(val);
p = p->next;
}
if(carry) p->next = new ListNode(1);
return head.next;
}
};