2. Add Two Numbers(Medium)

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.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

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类似于合并两个链表，处理的时候考虑进位即可。

//solution.cpp
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* p = new ListNode(0);
        auto l3 = p;
        auto pre = p;
        int temp;
        while(l1!=NULL && l2!=NULL){
            temp = l1->val+l2->val+l3->val;
            if (temp>=10){
                l3->val=temp-10;
                l3->next=new ListNode(1);
            }
            else{
                l3->val=temp;
                l3->next=new ListNode(0);
            }
            pre=l3;
            l3=l3->next;
            l1=l1->next;
            l2=l2->next;
        }
        while(l1!=NULL){
            temp=l1->val+l3->val;
            if (temp>=10){
                l3->val=temp-10;
                l3->next=new ListNode(1);
            }
            else{
                l3->val=temp;
                l3->next=new ListNode(0);
            }
            pre=l3;
            l3=l3->next;
            l1=l1->next;
        }
        while(l2!=NULL){
            temp=l2->val+l3->val;
            if (temp>=10){
                l3->val=temp-10;
                l3->next=new ListNode(1);
            }
            else{
                l3->val=temp;
                l3->next=new ListNode(0);
            }
            pre=l3;
            l3=l3->next;
            l2=l2->next;
    }
        if (l3->val==0)
            pre->next=NULL;
        return p;
    }
};