合并两个有序链表--实现1+2+3+....+n,时间复杂度为O（1）

1、合并两个有序链表，合并以后的链表依旧有效
C语言实现
链表结点定义如下：

typedef int DataType;typedef struct Node{    DataType _data;    struct Node* _pNext;}Node,*PNode;

实现代码:

void InitList(PNode* pHead) //初始化单链表{    assert(pHead);    //printf("%x\n",&pHead);    *pHead=NULL;}Node* BuyNode(DataType data)  //创建新结点{    Node *pNewNode=(Node*)malloc(sizeof(Node));    if(NULL!=pNewNode)    {        pNewNode->_data=data;         pNewNode->_pNext=NULL;    }    return pNewNode;}void PushBack(PNode* pHead,DataType data)  //插入新结点{    assert(pHead);    if(NULL==*pHead)    {        *pHead=BuyNode(data);    }    else    {        Node* pTailNode=*pHead;        while(pTailNode->_pNext)        {            pTailNode=pTailNode->_pNext ;        }        pTailNode->_pNext=BuyNode(data);    }}void PrintList(PNode pHead)  //打印单链表{    Node* pCurNode=pHead;    while(pCurNode)    {        printf("%d->",pCurNode->_data);         pCurNode=pCurNode->_pNext;    }    printf("NULL\n");}PNode MergeList(PNode pHead1, PNode pHead2)  //合并两个单链表{    PNode pL1=pHead1;    PNode pL2=pHead2;    PNode pNewHead=NULL;    PNode pTailNode=NULL;    if(pHead1==NULL)        return pHead2;    if(pHead2==NULL)        return pHead1;    pL1=pHead1;    pL2=pHead2;    if(pL1->_data <=pL2->_data)    {        pNewHead=pL1;        pTailNode=pL1;        pL1=pL1->_pNext;    }    else    {        pNewHead=pL2;        pTailNode=pL2;        pL2=pL2->_pNext;    }    while(pL1&&pL2)    {        if(pL1->_data<=pL2->_data)        {            pTailNode->_pNext=pL1;            pL1=pL1->_pNext;        }        else        {            pTailNode->_pNext=pL2;            pL2=pL2->_pNext;        }        pTailNode=pTailNode->_pNext;    }    if(NULL!=pL1)    {        pTailNode->_pNext =pL1;    }    if(NULL!=pL2)    {        pTailNode->_pNext=pL2;    }    return pNewHead;}void FunTest(){    Node *pHead1,*pHead2;    InitList(&pHead1);    PushBack(&pHead1,1);    PushBack(&pHead1,3);    PushBack(&pHead1,4);    PushBack(&pHead1,5);    PushBack(&pHead1,6);    PrintList(pHead1);    InitList(&pHead2);    PushBack(&pHead2,2);    PushBack(&pHead2,4);    PushBack(&pHead2,6);    PushBack(&pHead2,7);    PushBack(&pHead2,8);    PrintList(pHead2);    pHead1=MergeList(pHead1,pHead2);    PrintList(pHead1);}

链表合并以后的结果：

2、实现1+2+3+….+n,时间复杂度为O（1），要求不能使用乘除法，循环，条件判断，选择相关的关键字

 //1.利用构造函数求解--一次创建N个对象，将累加的算法放到构造函数中class Sort{public:     Sort()    {        N++;        Sum += N;    }     static void Reset()     {         N = 0;         Sum = 0;     }     static  int Get()     {         return Sum;     }private:    static  int N;    static  int Sum;};  int Sort::N = 0;  int Sort::Sum = 0; int Sum_Solution1(int n){    Sort::Reset();    Sort* a = new Sort[n];    delete a;    a = NULL;    return Sort::Get();}