数据结构：DHUOJ 单链表ADT模板应用算法设计：长整数加法运算（使用单链表存储计算结果）

单链表ADT模板应用算法设计：长整数加法运算（使用单链表存储计算结果）

时间限制: 1S类别: DS：线性表->线性表应用

题目描述：

输入范例：

-534564675768465476586798709880985345646757684654765867987098809853456467576846547658679870988098534564675768465476586798709880985345646757684654765867987098809853456467576846547658679870988098534564675768465476586798709880985345646757684654765867987098809853456467576846547658679870988098534564675768465476586798709880985345646757684654765867987098809853456467576846547658679870988098
435643754856985679435643754856985679435643754856985679435643754856985679435643754856985679435643754856985679435643754856985679435643754856985679435643754856985679435643754856985679435643754856985679435643754856985679435643754856985679435643754856985679

输出范例 :

-5345,6467,5768,4654,7658,6798,7098,8098,5345,6467,5768,4654,7658,6798,7098,8098,5345,6467,5768,4654,7658,6798,7098,8098,5345,6467,5768,4654,7658,6798,7098,8098,5345,6467,5768,4654,7658,6798,7098,8098,5345,6467,5768,4654,7658,6798,7098,8098,5345,6467,5768,4654,7658,6798,7098,8098,5345,6467,5768,4654,7658,6798,7098,8098,5345,6467,5768,4654,7658,6798,7098,8098,5345,6467,5768,4654,7658,6798,7098,8098,5345,6467,5768,4654,7658,6798,7098,8098,5345,6467,5768,4654,7658,6798,7098,8098
4356,4375,4856,9856,7943,5643,7548,5698,5679,4356,4375,4856,9856,7943,5643,7548,5698,5679,4356,4375,4856,9856,7943,5643,7548,5698,5679,4356,4375,4856,9856,7943,5643,7548,5698,5679,4356,4375,4856,9856,7943,5643,7548,5698,5679,4356,4375,4856,9856,7943,5643,7548,5698,5679,4356,4375,4856,9856,7943,5643,7548,5698,5679

-5345,6467,5768,4654,7658,6798,7098,8098,5345,6467,5768,4654,7658,6798,7098,8098,5345,6467,5768,4654,7658,6798,7098,8098,5345,6467,5768,4654,7658,6798,7098,8098,5345,2111,1392,9797,7801,8855,1455,0549,9647,0788,1412,0279,2801,6941,9155,2454,7797,0769,0089,0298,3283,1941,7242,0154,9701,8919,0069,8975,3302,2423,2241,8241,7402,0823,8219,8956,1979,2442,2723,3241,5488,8524,0124,7106,1960,1119,2742,3723,0488,6610,7824,9011,0110,1100,1419,3742,0970,1610,5911,6711,2014,9250,1400,2419

这里利用了几个关键函数：链表的逆置 比较两个链表的绝对值大小 链表的相加 输出链表

我的题解：

//DHUOJ 单链表ADT模板应用算法设计：长整数加法运算（使用单链表存储计算结果）
#include<iostream>
#include<cstring>using namespace std;
template<class ElemType>
struct Node
{
    ElemType data;
    Node<ElemType>* next;
    //构造函数1 用于node构造头结点
    Node(Node<ElemType>* ptr = NULL)
    {
        next = ptr;
    }
    //构造函数2 用于构造其他结点
    Node(const ElemType& item, Node<ElemType>* ptr = NULL)
    {
        next = ptr;
        data = item;
    }

public:
    ElemType getdata()
    {
        return data;
    }
    Node<ElemType>* getnext()
    {
        return next;
    }
    void set_next(Node<ElemType>* p)
    {
        this->next = p;
    }
    void set_date(ElemType num)
    {
        this->data = num;
    }
};
template<class ElemType>
class  LinkList {
private:
    Node<ElemType>* head;//头指针
    Node<ElemType>* tail;//尾指针

public:
    //无参构造函数
    LinkList()
    {
        head = new Node<ElemType>;
        head->next = NULL;
        tail = head->next;//头尾指针指向同一个内存
    }
    //有参构造
    LinkList(const ElemType& item)
    {
        head = new Node<ElemType>;
        tail = new Node<ElemType>;
        head->next = tail = new Node<ElemType>(item);//有参构造node
    }
    void display()
    {
        
       //避免出现以下情况 所以我们先判断是不是就一个单0
      if(head->getnext()->getdata()==0&&head->getnext()->getnext()==NULL)
         {
               cout<<"0";
                return ;
       }
         if (head->getdata() == -1)
            cout << "-";//如果是0就不能输出负号 
           //其实这样写也有问题 如果-0000 0005的就无法判断 所以我们上面解决了单0的情况
        Node<ElemType>* q = head->next;
        bool zeroflag = 0;
        bool firstflag = 0;
        //0要先特判 避免出现0 0000 0000的情况 也不行 会出现0 0000 0005 的情况无法输出
            while (q->next)
            {    
                if (q->getdata() == 0 && !zeroflag)
                {
                    q = q->next;
                    continue;
                }
                else if(q->getdata()!=0&&!zeroflag)
                {
                    zeroflag = 1;

                }
                if (!firstflag)
                {
                    cout << abs(q->data) << ",";
                    firstflag = 1;
                }
                else
                    printf("%04d,", abs(q->data));
                q = q->next;
            
            }
        if (q == head->next)//只有一位特判
            cout << abs(q->data);
        else
        {
            if (!firstflag)
            {
                cout << abs(q->data);

            }
            else
            {
                if (!zeroflag)
                {
                    cout << abs(q->data);

                }
                else
                {
                    printf("%04d", abs(q->data));
                }
                
            }
                
        }
            
        cout << endl;
    }
    int size()
    {
        if (head->next == NULL)
            return 0;
        int len = 0;
        Node<ElemType>* q;
        q = head->next;
        while (q)
        {
            len++;
            q = q->next;
        }
        return len;
    }
    void push_back(ElemType x)
    {
        Node<ElemType>* q = new Node<ElemType>;//新建结点
        q->data = x;
        q->next = NULL;
        if (size() == 0)
        {
            head->next = q;

        }
        else
        {
            tail->next = q;

        }
        tail = q;

    }
    Node<ElemType>* get_front(Node<ElemType>* p)
    {//获取一个指针的上一个，头指针和非法指针会报错.
        Node<ElemType>* q;
        q = head->next;
        while (q->next != NULL)
        {
            if (q->next == p)
                return q;
            q = q->next;
        }
    }
    Node<ElemType>* get_next(Node<ElemType>* p)
    {
        return p->next;
    }
    Node<ElemType>* get_address(int i)//获取指定下标的地址
    {
        Node<ElemType>* q;
        q = head->next;
        while (i--)
            q = q->next;
        return q;
    }
    ElemType at(int i)
    {
        Node<ElemType>* q = get_address(i);
        return q->data;
    }
    bool del_p(Node<ElemType>* p)//传入一个指针 删除
    {
        if (size() == 0)
            return false;
        if (tail->next == NULL)//如果只有一个元素
        {
            if (p == tail)//如果这个指针是那个唯一的指针
            {
                delete tail;
                tail = NULL;
                return true;
            }
            else
                return false;//如果不是

        }
        if (p == tail)//如果删除的是尾指针 
        {
            Node<ElemType>* q = get_front(p);
            q->next = NULL;
            tail = q;
            delete p;
            return true;
        }
        //其他的
        Node<ElemType>* q = get_front(p);
        q->next = p->next;
        delete p;
        return true;

    }
    bool del(int i)//删除指定位置的元素
    {
        return del_p(get_address(i));
    }
    Node<ElemType>* get_head()
    {
        return head;
    }
    Node<ElemType>* get_tail()
    {
        return tail;
    }
    void set_head(Node<ElemType>* p)
    {

        head = p;
    }
    void set_tail(Node<ElemType>* p)
    {
        tail = p;
    }
    void ListReverse()
    {
        Node<ElemType>* a, * b, * temp;
        a = head->getnext();
        ElemType datas = head->getdata();
        temp = NULL;
        b = NULL;
        bool flag = 0;//设置尾指针的标志
        while (a)
        {
            b = a;
            a = a->getnext();
            b->set_next(temp);
            if (!flag)
            {
                tail = b;
                flag = 1;
            }
            temp = b;
        }
        Node<ElemType>* newhead = new Node<ElemType>;
        newhead->set_next(b);
        newhead->set_date(datas);
        head = newhead;

    }
};


//从长整数的低位开始拆分（4位为一组，即不超过9999的非负整数），依次存放在单链表的每个结点的数据域中；
//头结点的数据域存放正负数标志（正数或0：1，负数：-1）。
template<class ElemType>
void Input_Int_Division(LinkList<ElemType>& L, string& str, int& length)
{
    Node<ElemType>* head = L.get_head();
    bool zhengfu_flag = 0;//记录正负的flag
    int l = str.length();
    if (str[0] == "-")//先特判正负数    
    {

        zhengfu_flag = 1;
        head->set_date(-1);

    }
    else
    {
        head->set_date(1);
    }
    int i = 0;
    if (zhengfu_flag)
        i = 1;
    int t0 = l % 4;
    if (t0 == 0)
        t0 = 4;
    int t = 0;//记录位数
    int num = 0;//存四位数
    bool changeflag = 0;//判断标志 判断是否有进行第一次
    for (; i < t0; ++i)
    {
        num = num * 10 + (str[i] - "0");
        changeflag = 1;
    }
    if (changeflag)
    {
        length++;//记录长度
        L.push_back(num);
        num = 0;
    }

    for (; i < str.length(); ++i)
    {

        num = num * 10 + (str[i] - "0");
        t++;
        if (t == 4)
        {
            length++;//记录长度
            L.push_back(num);
            t = 0;
            num = 0;
        }
    }
    //L.display();
}
//两个长整数的 绝对值 大小比较
//（x>y 返回值为1；x<y 返回值为2；x=y 返回值为0；）
template<class ElemType>
int Two_LongNum_Compare(LinkList<ElemType>& A, LinkList<ElemType>& B, const int& len_A, const int& len_B)
{
    Node<ElemType>* head1 = A.get_head();
    Node<ElemType>* head2 = B.get_head();

    //正数的情况 先看长度 
    if (len_A > len_B)
    {
        return 1;
    }
    else if (len_A < len_B)
    {
        return 2;
    }
    else if (len_A == len_B)
    {
        Node<ElemType>* a, * b;
        a = head1->getnext();
        b = head2->getnext();
        while (a)
        {
            if (a->getdata() > b->getdata())
                return 1;
            else if (a->getdata() < b->getdata())
                return 2;
            else
                a = a->getnext(), b = b->getnext();
        }
        return 0;
    }


}
template<class ElemType>
void swaps(LinkList<ElemType>& a, LinkList<ElemType>& b)
{
    LinkList<ElemType>temp = a;
    a = b;
    b = temp;
}
template<class ElemType>
void Listadds(LinkList<ElemType>& a, LinkList<ElemType>& b, int& la, int& lb)
{
    Node<ElemType>* x, * y;

    int ans = 0;
    if (la < lb)
    {
        //swap一下两个
        swaps(a, b);
        int temp = la;
        la = lb;
        lb = temp;
    }
    x = a.get_head()->getnext();
    y = b.get_head()->getnext();//两个指针的创建必须放在 交换判断之后
    int m = a.get_head()->getdata();//m n 存储符号
    int n = b.get_head()->getdata();//存储符号也必须放在交换判断之后

    //第一步 判断结果的最高位//!必须再加法前进行 ！！因为a会随着加法而改变 引用传递
    bool flag = 0;//标记元素
    int comp = Two_LongNum_Compare(a, b, la, lb);


    while (y)//先把每一位结点的数值加起来
    {
        ans = x->getdata() * m + y->getdata() * n;
        x->set_date(ans);
        x = x->getnext();
        y = y->getnext();
    }
    //把长的位都化成同号的 不然接下来进位会出问题
    while (x)
    {
        x->set_date(x->getdata() * m);
        x = x->getnext();
    }
    //再做进位处理
    
    if (comp == 1)
    {
        flag = 1;
    }
    //第二步 开始逐位遍历
    x = a.get_head()->getnext();
    int zheng_fu = 0;//判断正负的符号
    if (flag)
        zheng_fu = a.get_head()->getdata();
    else
        zheng_fu = b.get_head()->getdata();
    if (zheng_fu > 0)//分正负两种情况 先看是正的    
    {
        a.get_head()->set_date(1);//定最高位符号
        while (x->getnext())//最后一个结点先不遍历 最后单独讨论
        {
            if (x->getdata() > 9999)
            {
                x->set_date(x->getdata() - 10000);
                x->getnext()->set_date(x->getnext()->getdata() + 1);
                //下一位就加一
            }
            else if (x->getdata() < 0)
            {
                x->set_date(x->getdata() + 10000);
                x->getnext()->set_date(x->getnext()->getdata() - 1);
            }
            x = x->getnext();
        }
        //讨论最后一位 是否要进位
        if (x->getdata() > 9999)
        {
            x->set_date(x->getdata() - 10000);
            a.push_back(1);
        }
    }
    else //讨论负数的情况
    {
        a.get_head()->set_date(-1);//定最高位符号
        while (x->getnext())//最后一个结点先不遍历 最后单独讨论
        {
            if (x->getdata() < -9999)
            {
                x->set_date(x->getdata() + 10000);
                x->getnext()->set_date(x->getnext()->getdata() - 1);
                //下一位就加一
            }
            else if (x->getdata() > 0)
            {
                x->set_date(x->getdata() - 10000);
                x->getnext()->set_date(x->getnext()->getdata() + 1);
            }
            x = x->getnext();
        }
        //讨论最后一位 是否要进位
        if (x->getdata() < -9999)
        {
            x->set_date(x->getdata() + 10000);
            a.push_back(1);
        }
    }
}

int main()
{
    LinkList<int>head1, head2;
    string str1, str2;
    getline(cin, str1);
    getline(cin, str2);
    int la = str1.length();
    int lb = str2.length();
    if (str1[0] == "-")
        la -= 1;
    if (str2[0] == "-")
        lb -= 1;
    int length1 = 0;
    int length2 = 0;
    Input_Int_Division(head1, str1, length1);
    Input_Int_Division(head2, str2, length2);
    head1.display();
    head2.display();
    cout << endl;
    head1.ListReverse();
    head2.ListReverse();
    Listadds(head1, head2, la, lb);
    head1.ListReverse();
    head1.display();
    //swaps(head1, head2);
    //head1.display();
    //head2.display();
    //cout << endl;
    //int comp = Two_LongNum_Compare(head1, head2, length1, length2);
    //cout << comp;
    //cout << length<<endl;
    //head.ListReverse();
    //head.display();

    return 0;

}