合并两个有序的链表和计算1+2+3+4....

//合并两个有序链表，使合并后依然有序PNode MergeList(PNode pHead1, PNode pHead2)   //递归实现{    if (pHead1 == NULL)        return pHead2;    else if (pHead2 == NULL)        return pHead1;    PNode newhead = NULL;    if (pHead1->data < pHead2->data)    {        newhead = pHead1;        newhead->_next = MergeList(pHead1->_next, pHead2);    }    else    {        newhead = pHead2;        newhead->_next = MergeList(pHead1, pHead2->_next);    }    return newhead;}/*##############################################################1+2+3+4+5+........+n###############################################################*/n*(n - 2) / 2num = 0;while (n--){    mun = num + n;}//利用构造函数  我们可以先定义一个类型，然后创建n个该类型的实例，//这样构造函数会被调用n次，可以把累加代码放在构造函数中class GetAddnum{public:    GetAddnum()    {        N++;        sum = sum + N;    }    static void Reset() //初始化    {        N = 0;        sum = 0;    }    static unsigned int printfnum()    {        return sum;    }private:    static unsigned int N;    //静态变量函数    static unsigned int sum;};unsigned int GetAddnum_solution(int n){    GetAddnum::Reset();  //初始化    GetAddnum* p = new GetAddnum[n];    delete[] p;    p = NULL;    return GetAddnum::printfnum();}int main(){    GetAddnum_solution(7);}