[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;     }};