Add Two Numbers

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;
    }
};