2. Add Two Numbers

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.

Example

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

注意补齐队列，高位在链尾，处理进位

/**
 * 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 *res = new ListNode(-1);
        ListNode *cur = res;
        int carry=0;
        while(l1 || l2)
        {
            int n1 = l1 ? l1 -> val : 0;
            int n2 = l2 ? l2 -> val : 0;//以0补齐
            int sum = n1 + n2 + carry;
            carry=sum/10;
            cur->next =new ListNode(sum%10);
            cur=cur->next;
            if(l1)
                l1=l1->next;
            if(l2)
                l2=l2->next;
        }
        if(carry)
            cur->next=new ListNode(carry);//处理最高位的进位
        return res->next;
    }
};