Subtraction of large numbers using linked list for C + +

1. Preface

The implementation of the large number of subtraction is the winter vacation in C + + 's big job, originally I was written in string, but later to see the job requirements to use the chain list to achieve, so I do not want to use the list of the realization of the

2.Num class 2.1Node Class

First, the internal node class is used to create a one-way list, and size is used to calculate sizes to facilitate comparisons between NUM objects

    class Node    {    public:        int val;        Node*next;        Node(int v, Node*n)        {            val = v;            next = n;        }    };    Node*head;    int size;

2.2 Constructors, assignment functions, and destructors

The first is the default constructor, which may not actually be used, but for convenience, it is still written

    Num()    {        head = NULL;        0;        false;    }

Then is the primary constructor, when encountering '. ' , we use-100来 as a marker.

The linked list is a number of reverse construction, convenient for adding and reducing operations

    Num(const string&s)    {        false;        head = NULL;        0;        for (int0; i < s.size(); i++)        {            if','||s[i]=='-')                continue;            size++;            if'.')                new Node(-100, head);            else                new'0', head);        }    }

Next is the copy constructor and the assignment function, which are basically the same

It is important to note that the assignment function first confirms whether the assignment object is itself, and if it is itself, returns the value of the

Num (Constnum& num) {D = num.        D        size = Num.size;        head = NULL; Node*cur = Num.head, *temp=head; while(cur) {if(!head) {head =NewNode (Cur->val, NULL);            temp = head; }Else{Temp->next =NewNode (Cur->val, NULL);            temp = temp->next;        } cur = cur->next; }} num&operator=(ConstNum&num) {if( This= = &num)return* This;        Free ();        size = Num.size;        head = NULL; D = num.        D Node*cur = Num.head, *temp=head; while(cur) {if(!head) {head =NewNode (Cur->val, NULL);            temp = head; }Else{Temp->next =NewNode (Cur->val, NULL);            temp = temp->next;        } cur = cur->next; }return* This; }

Finally, the destructor, we only need to reclaim the memory of the linked list

To facilitate recycling of the linked list memory within the function, write the recycle operation as the free function

    ~Num()    {        free();    }    void free()    {        0;        while (head)        {            Node*temp = head->next;            delete head;            head = temp;        }        head = NULL;    }

The above is the basic constituent function of the Num class

2.3 Addition operation and subtraction operation

For ease of operation, the addition and subtraction of the large number in front, the decimal in the following, directly to the object itself, so use +=,-= and return the object itself reference

Add and subtract operations nothing to say, starting from the lowest bit to add in turn

An operation that determines whether a decimal point is performed when rounding or borrow, in case of an incorrect operation

num&operator+=(ConstNum&num) {Node*pa = head, *PB = Num.head; while(pa| | PB) {intb =0;if(PB)                {b = pb->val;            PB = pb->next; }if(Pa->val! =-100) {Pa->val + = b;if(Pa->val >9)                {if(Pa->next) {if(Pa->next->val = =-100) pa->next->next->val++;Elsepa->next->val++; }Else{Pa->next =NewNode (1, NULL);                    size++; } Pa->val-=Ten;        }} PA = pa->next; }return* This; } num&operator-=(ConstNum&num) {Node*pa = head, *PB = Num.head; while(PA | | pb) {intb =0;if(PB)                {b = pb->val;            PB = pb->next; }if(Pa->val! =-100) {Pa->val-= b;if(Pa->val <0)                {if(Pa->next->val = =-100) pa->next->next->val--;Elsepa->next->val--; Pa->val + =Ten;        }} PA = pa->next; }return* This; }

2.4 Multiplication operation

Multiplication is actually like a column calculation, which involves rounding, but it's easy to understand

    operator*(const Num&num)    {        '0'));        Node *pr = res.head, *temp;        for (Node*pa = head; pa != NULL; pa = pa->next)        {            int0;            temp = pr;            for (Node*pb = num.head; pb != NULL; pb = pb->next,pr=pr->next)            {                int temp = pa->val*pb->val + carry + pr->val;                10;                10;            }            pr->val += carry;            pr = temp->next;        }        return res;    }

2.5 Division Operation

Division operation we use the divisor minus removal number multiplied by a dry 10

Like 58 divided by 5 first with 58-50=8 then quotient plus 10 then with 8-5=3 quotient plus 1 because 3:5 small, do not operate the final result is 11

However, the topic requires division operation to keep 10 decimal places rounding, we only need to dividend times the 10^11, and finally add the decimal point to rounding operations

We first implemented two functions related to 10

    void Mul10(constint& x)    {        for (int0; i < x; ++i)        {            new Node(0, head);            size++;        }    }    void Div10(constint&x)    {        for (int0; i < x; ++i)        {            Node*temp = head->next;            delete head;            head = temp;            size--;        }    }

Then the division operation, RES is used to return the result, p is used to add to res

D for subtracting from dividend

Numoperator/(Constnum&num) {num res (string (1,' 0 '), p (String (1,' 1 ')); Res. Btrue; P.D =true; num d = num; while(num<=* This)        {intlen = size-num.size;            D.mul10 (len); P.mul10 (len);intz =0;if(* This< D) {D.DIV10 (1); P.DIV10 (1); z =-1; }if(Res.head->val = =0&&res.size==1) res = p;Elseres + = P; * This-= D;            Balance ();            D.DIV10 (len + z);        P.DIV10 (len + z); }returnRes }

We notice that there is a comparison < operation in the division operation, and then the overload < function

Subtraction of large numbers using linked list for C + +