LeetCode 2 Add Two Numbers

原题：

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.

中文：

给你两个链表，现在让你计算两个链表的和，每个链表的节点保存个位数，进位运算累加至下一个节点。

struct ListNode
{
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2)
    {
        if(l1==NULL)
            return l2;
        if(l2==NULL)
            return l1;

        ListNode *head=new ListNode(0);
        ListNode *p=head;
        int pre=0;

        head->val=(l1->val+l2->val)%10;
        pre = (l1->val+l2->val)/10;
        l1=l1->next;
        l2=l2->next;
        while(l1!=NULL||l2!=NULL)
        {
            ListNode *tmp=new ListNode(pre);
            if(l1!=NULL)
                tmp->val+=l1->val;
            if(l2!=NULL)
                tmp->val+=l2->val;


            pre=(tmp->val)/10;
            tmp->val=(tmp->val)%10;
            p->next=tmp;
            p=tmp;
            if(l1!=NULL)
                l1=l1->next;
            if(l2!=NULL)
                l2=l2->next;
        }
        if(pre!=0)
        {
            ListNode *tmp=new ListNode(pre);
            p->next=tmp;
            p=tmp;
        }
        return head;
    }
};

解答：

好久没做LeetCode了，也好久没写链表了，其实思路很简单，就是以此加起来，保存进位相即可。