标签：des its ati bsp tno exce sel inpu res
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
思路：新建一个新的链表，头指针指向一个0，但实际从头指针的下一个开始指向计算的结点，当l1，l2其中一个不为空或者carry位不为0，就进行计算，直到条件不成立。
/** * 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) { int carry=0; ListNode *p=new ListNode(0); ListNode *q=p; while(l1||l2||carry) { if(l1) { carry+=l1->val; l1=l1->next; } if(l2) { carry+=l2->val; l2=l2->next; } p->next=new ListNode(carry%10); carry/=10; p=p->next; } return q->next; } };
标签：des its ati bsp tno exce sel inpu res
原文地址：http://www.cnblogs.com/wsw-seu/p/7642899.html
