2. Add Two Numbers

题目：

解答：

常规的链表数值求和方式。

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) {
        ListNode* head = new ListNode(-1), *p = head;
        int c = 0;
        while(l1 || l2){
            int num1 = l1 ? l1->val : 0;
            int num2 = l2 ? l2->val : 0;
            int val = (c + num1 + num2) % 10;
            c = (c + num1 + num2) / 10;
            ListNode* node = new ListNode(val);
            p->next = node;p = node;
            l1 = l1 ? l1->next : l1;
            l2 = l2 ? l2->next : l2;
        }
        if(c > 0){
            ListNode* node = new ListNode(c);
            p->next = node;
        }
        p = head->next;
        delete head;
        return p;
    }
};