【Leetcode】2.Add Two Numbers 解题

【题目描述】：
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

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【思路】：
以l1为基准，直接对其进行修改，避免了使用新链，new出很多个新节点。

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【我的代码】：
1556 / 1556 test cases passed.
Runtime: 36 ms
Your runtime beats 63.70% of cppsubmissions.

/** * 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 *result = l1;//记录链首        ListNode *p = l1;//跟踪现需处理的节点        ListNode *tmpNode = new ListNode(0);//进位数据节点，临时存储        //对l1 l2并行处理        while (l1 && l2) {            l1->val = l1->val + l2->val + tmpNode->val;            tmpNode->val = l1->val / 10;            l1->val = l1->val % 10;            p = l1;            l1 = l1->next;            l2 = l2->next;        }        //l1 l2不同长度，l1 l2同一长度则可以跳过        ListNode *pre_p = p;        if ((l1 && !l2) || (!l1 && l2)) {            //l2长，l1长的话，p不变            if (!l1 && l2) {                p->next = l2;            }            p = p->next;            while (p) {                p->val = p->val + tmpNode->val;                tmpNode->val = p->val / 10;                p->val = p->val % 10;                pre_p = p;                p = p->next;            }        }        //如果最后还有进位，则新new一个节点        if (tmpNode->val != 0) {            ListNode *q = new ListNode(tmpNode->val);            pre_p->next = q;        }        return result;    }};