leetcode 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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
我的思路：把所有的数据往l1上拼接，如果l1比较短，那么就往上面拼接。

/**
 * 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* p1 = l1;
        ListNode* p2 = l2;
        ListNode* p3 = NULL;
        ListNode* p4 = NULL;
        int mark = 0;
        while (p1 != NULL && p2 != NULL) {
            p1->val = p1->val + p2->val;
            if (mark) p1->val++, mark = 0;
            if (p1->val >= 10) p1->val = p1->val % 10, mark = 1;
            p3 = p1;
            p1 = p1->next;
            p2 = p2->next;
        }
        if (p2 != NULL) {
            while (p2 != NULL) {
                ListNode *x = new ListNode(p2->val);
                p3->next = x;
                if (mark) p3->next->val++, mark = 0;
                if (p3->next->val >= 10) p3->next->val = p3->next->val % 10, mark = 1;
                p2 = p2->next;
                p3 = p3->next;
            }
            if (mark) {
                ListNode *x = new ListNode(1);
                p3->next = x;
                p3 = p3->next;
            }
        }
        else if (p1 != NULL) {
            while (p1 != NULL) {
                if (mark) p1->val++, mark = 0;
                if (p1->val >= 10) p1->val = p1->val % 10, mark = 1;
                p4 = p1;
                p1 = p1->next;
            }
            if (mark) {
                ListNode *x = new ListNode(1);
                p4->next = x;
                p4 = p4->next;
            }
        } 
        else {
            if (mark) {
                ListNode *x = new ListNode(1);
                p3->next = x;
                p3 = p3->next;
            }
        }
        return l1;
    }
};

虽然过了，但是看了别人的代码后感觉自己的代码太长了，差距都是对比出来的

/**
 * 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* header = new ListNode(0);
        ListNode* now = header;
        while(l1 || l2 || carry){
            now->next = new ListNode(0);
            now = now->next;
            now->val = (l1?l1->val:0) + (l2?l2->val:0) + carry;
            carry = now->val / 10;
            now->val = now->val%10;
            l1=l1?l1->next:nullptr;
            l2=l2?l2->next:nullptr;
        }
        ListNode* out = header->next;
        delete header;
        return out;
    }
};

今天是开学的第一天，加油。