21-Add Two Numbers-Leetcode

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
思路：从左边对齐，开始相加，将产生的进位置为下一个指针的初始值，这里主要需要考虑到几种情况，两个链表相等或不等两类，
最重要的是根据进位决定next指针是否为空或者是一个新节点。

/** * 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 * head = new ListNode(0);
        ListNode * beg = head;
        int pre=0;
        while(l1!=NULL||l2!=NULL){
            if(l1!=NULL&&l2!=NULL){
                head->val = head->val+l1->val+l2->val;
                pre = head->val/10;
                head->val = (head->val)%10;
                l1 = l1->next;
                l2 = l2->next;
                if(l1==NULL&&l2==NULL&&pre==0)head->next = NULL;//这个if很重要
                else head->next = new ListNode(pre);
                head = head->next;
            }
            else if(l1!=NULL&&l2==NULL){
                head->val = head->val+l1->val;
                pre = head->val/10;
                head->val = (head->val)%10;
                l1 = l1->next;
                if(l1==NULL&&pre==0)head->next = NULL;
                else head->next = new ListNode(pre);
                head = head->next;
            }
            else if(l1==NULL&&l2!=NULL)
            {
                head->val = head->val+l2->val;
                pre = head->val/10;
                head->val = (head->val)%10;
                l2 = l2->next;
                if(l2==NULL&&pre==0)head->next = NULL;
                else head->next = new ListNode(pre);
                head = head->next;
            }
            else{
                head->val = pre;
                return beg;
            }
        }
        return beg;
    }
};