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

/**
 * 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,*p2=l2;
        ListNode* result=NULL,*temp=NULL,*presult=NULL;//result為結果連結串列頭，temp為每位上相加後的值，presult為將temp連結到result上
        int up=0;
        while(p1!=NULL&&p2!=NULL){
                temp= new ListNode(p1->val+p2->val);
                temp->val=up+temp->val;//先加上進位得到本位之和
                up=temp->val/10; //得到本位之和的進位值
                temp->val=temp->val%10;//得到本位值
                if(result==NULL){//空連結串列第一次賦值
                    result=temp;
                    presult=result;
                }
                else{
                    presult->next=temp;
                    presult=temp;
                }
                p1=p1->next;
                p2=p2->next;
                
            }
        while(p2!=NULL){
            temp= new ListNode(p2->val);
            temp->val=up+temp->val;
            up=temp->val/10; 
            temp->val=temp->val%10;
            if(result==NULL){
                    result=temp;
                    presult=result;
                }
            else{
                    presult->next=temp;
                    presult=temp;
            }
            p2=p2->next;
        }
        while(p1!=NULL){
            temp= new ListNode(p1->val);
            temp->val=up+temp->val;
            up=temp->val/10; 
            temp->val=temp->val%10;
            if(result==NULL){
                    result=temp;
                    presult=result;
                }
            else{
                    presult->next=temp;
                    presult=temp;
            }
            p1=p1->next;
        }
        if(up>0)//[5] [5]結果為[0][1]而不是[0]
        {
            temp= new ListNode(up);
            presult->next=temp;
        }
    return result;
    }
};