221.假定用一個連結串列表示兩個數，其中每個節點僅包含一個數字。假設這兩個數的數字順序排列，請設計一種方法將兩個數相加，並將其結果表現為連結串列的形式。
對於這道題，可以先把兩個連結串列翻轉，再依次相加，再把相加的結果再次翻轉。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */


class Solution {
public:
    /*
     * @param l1: The first list.
     * @param l2: The second list.
     * @return: the sum list of l1 and l2.
     */
 

    ListNode * addLists2(ListNode * l1, ListNode * l2) {
        if(!l1&&!l2)
            return NULL;

        ListNode *head = new ListNode;
        ListNode *tail = head;
        ListNode *p1 = reverse(l1);//翻轉連結串列l1
        ListNode *p2 = reverse(l2);//翻轉連結串列l2
        head->next = NULL;
        int
 
 val1,val2;
        int flag = 0;
        while(p1||p2)//翻轉後的連結串列元素依次相加
        {
            if(p1)
            {
                val1 = p1->val;
                p1 = p1->next;
            }
            else
                val1 = 0;
            if(p2)
            {
                val2 = p2->val;
                p2 = p2->next
 
;
            }
            else
                val2 = 0;
            int temp = flag+val1+val2;
            flag = temp/10;
            temp = temp%10;
            ListNode *pTest = new ListNode;
            pTest->val = temp;
            pTest->next = NULL;
            tail->next = pTest;
            tail = tail->next;
        }
        if(flag)
        {
            ListNode *pTest = new ListNode;
            pTest->val = flag;
            pTest->next = NULL;
            tail->next = pTest;
        }
        tail = head;
        head = head->next;
        delete tail;
        return reverse(head);//翻轉計算結果
    }
    ListNode * reverse(ListNode *head)//翻轉連結串列
    {
        ListNode *p = new ListNode;
        ListNode *temp = NULL;
        p->next = NULL;
        temp = head;
        head = p;
        p = temp;
        while(p)
        {
            temp = p;
            p = p->next;
            temp->next = head->next;
            head->next = temp;
        }
        temp = head;
        head = head->next;
        return head;
    }
};