Leetcode 02 Add Two Numbers

Add Two Numbers

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

链表处理

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2)
    {
        ListNode* result_head = new ListNode(0);
        ListNode* p;
        p = result_head;
        while (l1 != NULL&&l2 != NULL)
        {
            ListNode* tmp = new ListNode(0);
            (*p).val += (*l1).val + (*l2).val;
            (*p).next = tmp;
            if ((*p).val >= 10)
            {
                (*p).val %= 10;
                (*tmp).val += 1;
            }
            p = tmp;
            l1 = (*l1).next;
            l2 = (*l2).next;
        }
        if (l1 == NULL&&l2 == NULL) {}
        else
        {
            if (l1 == NULL&&l2 != NULL)
            {
                while (l2 != NULL)
                {
                    (*p).val += (*l2).val;
                    ListNode* tmp = new ListNode(0);
                    (*p).next = tmp;
                    if ((*p).val >= 10)
                    {
                        (*p).val %= 10;
                        (*tmp).val += 1;
                    }
                    p = tmp;
                    l2 = (*l2).next;
                }
            }
            else
            {
                if (l1 != NULL&&l2 == NULL)
                {
                    while (l1 != NULL)
                    {
                        (*p).val += (*l1).val;
                        ListNode* tmp = new ListNode(0);
                        (*p).next = tmp;
                        if ((*p).val >= 10)
                        {
                            (*p).val %= 10;
                            (*tmp).val += 1;
                        }
                        p = tmp;
                        l1 = (*l1).next;
                    }
                }
            }
        }
        p = result_head;
        while (((*p).next)->next != NULL)  p = (*p).next;
        if (((*p).next)->val == 0)
        {
            delete (*p).next;
            (*p).next = NULL;
        }
        return result_head;
    }
};