#2. Add Two Numbers

题目描述：

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

/** * 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 digit_Num,extra_Num=0,random;        ListNode *tou=NULL,*p1,*p2,*temp;        while(l1!=NULL&&l2!=NULL){            digit_Num = (l1->val+l2->val+extra_Num)%10;            //cout<<"digit_Num = "<<digit_Num<<endl;            if(tou==NULL){                //new malloc还是有相似的地方                //p2 = p1=tou = new ListNode(digit_Num+extra_Num);                p2 = p1 = tou =(ListNode*)new ListNode(digit_Num+extra_Num);                //cout<<"tou "<<tou->val<<endl;            }else{                p1 = (ListNode*)new ListNode(digit_Num);                p2->next = p1;                p2 = p1;            }            extra_Num = (l1->val+l2->val+extra_Num)/10;            //cout<<"extra_Num = "<<extra_Num<<endl;            //temp = l1;            l1 = l1->next;            //free(temp);            //temp = l2; 可能是栈上的内存 不能瞎释放            //free(temp);            l2 = l2->next;        }        //cout<<" Here extra_Num = "<<extra_Num<<endl;        if(l1==NULL&&l2==NULL){            //cout<<"3  code here"<<"ex = "<<extra_Num<<endl;            if(extra_Num!=0)                p2->next = (ListNode*)new ListNode(extra_Num);        }else if(l1==NULL){            temp = l2;            while(temp!=NULL&&extra_Num!=0)            {                random = temp->val;                //cout<<"temp->val = "<<temp->val<<endl;                temp->val = (extra_Num + temp->val)%10;            //  cout<<"temp->val = "<<temp->val<<endl;                extra_Num = (extra_Num+random)/10;                temp = temp->next;            }            //cout<<"ex = "<<extra_Num<<endl;            if(extra_Num!=0){                temp = l2;                //cout<<"1 Code come to here"<<endl;                while(temp->next!=NULL)                    temp = temp->next;                temp->next = (ListNode*)new ListNode(extra_Num);            }            p2->next = l2;        }else if(l2==NULL){            temp = l1;            while(temp!=NULL&&extra_Num!=0)            {                random = temp->val;                temp->val = (extra_Num + temp->val)%10;                extra_Num = (extra_Num+random)/10;                temp = temp->next;            }        //  cout<<"ex = "<<extra_Num<<endl;            if(extra_Num!=0){                temp = l1;            //  cout<<"2 Code come to here"<<endl;                while(temp->next!=NULL)                    temp = temp->next;                temp->next = (ListNode*)new ListNode(extra_Num);            }            p2->next = l1;        }        return tou;    }};

无限大的数字相加问题，一般会模拟加法运算，同时利用栈来提高效率，本题是一样的，倒序存放的链表和栈是一样的，操作复杂了一点，以上代码可以做一点优化，但是算法本身没有办法提速。

能不动手尽量不动手，在大脑中模拟算法的计算步骤和流程，最好在脑子中写好之后再动手，否则无法提高，只是根据OJ判定的数据来填漏洞而已，太笨了。