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：
sum=(s1+s2+carry)%10;
carry=(s1+s2+carry)/10;
由以上关系即可得到code；
code:
/** * 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 head(-1); int carry=0; ListNode *cur=&head; for(ListNode *p1=l1,*p2=l2;p1!=NULL||p2!=NULL; p1=p1==NULL ? NULL:p1->next,p2=p2==NULL ? NULL:p2->next, cur=cur->next) { const int s1=p1==NULL ? 0:p1->val; const int s2=p2==NULL ? 0:p2->val; const int sum=(s1+s2+carry)%10; carry=(s1+s2+carry)/10; cur->next=new ListNode(sum); } if(carry>0) cur->next=new ListNode(carry); return head.next; } };
原文：http://www.cnblogs.com/chengyuz/p/4037300.html
