leetcode练习 Add Two Numbers

感觉很久没有接触链表，又打算在处理图的时候使用邻接表
稍微做一道小题目熟悉一下
很简单，权当练手，重点是后面图的相关问题

/** * 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) {        if (l1 == NULL) return l2;        if (l2 == NULL) return l1;        ListNode* result = new ListNode(0);        ListNode* tail = new ListNode(0);        int flag = 0;        result->next = tail;        tail->val = (l1->val+l2->val)%10;        if ((l1->val+l2->val) >= 10) {            flag = 1;        } else {            flag = 0;        }        ListNode* temp1 = l1->next;        ListNode* temp2 = l2->next;        while (temp1 != NULL && temp2 != NULL) {            ListNode* n = new ListNode((temp1->val+temp2->val+flag)%10);            tail->next = n;            tail = n;            if ((temp1->val+temp2->val+flag) >= 10) {                flag = 1;            } else {                flag = 0;                }            temp1 = temp1->next;            temp2 = temp2->next;        }        if (!temp1) {            while (temp2) {                ListNode* n = new ListNode((temp2->val+flag)%10);                if ((temp2->val+flag) >= 10) {                    flag = 1;                } else {                    flag = 0;                }                tail->next = n;                tail = n;                temp2 = temp2->next;            }        }        else if (!temp2) {            while (temp1) {                ListNode* n = new ListNode((temp1->val+flag)%10);                if ((temp1->val+flag) >= 10) {                    flag = 1;                } else {                    flag = 0;                }                tail->next = n;                tail = n;                temp1 = temp1->next;            }        }        if (flag) {            ListNode* f = new ListNode(flag);            tail->next = f;            tail = f;        }        result = result->next;        return result;    }};