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所以我的作业要求我们使用双向链表将数字相加或相乘并打印出来。我能够让它适用于整数,但我无法弄清楚要更改什么以使其也适用于小数。这是我到目前为止所得到的。我知道这不是最有效或最干净的代码,但如果它对你没有意义,我可以尝试澄清一些事情例如,如果我执行 50382+9281 或 482891*29734,此程序将正常工作,但我需要让它在 4.9171+49.2917 或 423.135*59 之类的情况下工作
编辑:假设 int 值是 double 值。我在我的实际代码中更改了它,但是当我进行数学计算时的结果仍然给我一个整数,所以我需要弄清楚如何在正确的位置插入小数
#include <iostream>
#include <fstream>
#include <string>
#include <stdio.h>
#include <cstdlib>
#include <cstring>
using namespace std;
// A recursive program to add two linked lists
#include <stdlib.h>
#include <assert.h>
#include <math.h>
#include <string.h>
// A linked List Node
struct node
{
int data;
node* next;
node *prev;
};
typedef struct node node;
class LinkedList{
// public member
public:
// constructor
LinkedList(){
int length = 0;
head = NULL; // set head to NULL
node *n = new node;
n->data = -1;
n->prev = NULL;
head = n;
tail = n;
}
// This prepends a new value at the beginning of the list
void addValue(int val){
node *n = new node(); // create new Node
n->data = val; // set value
n->prev = tail; // make the node point to the next node.
// head->next = n;
// head = n;
// tail->next = n; // If the list is empty, this is NULL, so the end of the list --> OK
tail = n; // last but not least, make the head point at the new node.
}
void PrintForward(){
node* temp = head;
while(temp->next != NULL){
cout << temp->data;
temp = temp->next;
}
cout << '\n';
}
void PrintReverse(){
node* temp = tail;
while(temp->prev != NULL){
cout << temp->data;
temp = temp->prev;
}
cout << '\n';
}
void PrintReverse(node* in){
node* temp = in;
if(temp->prev== NULL){
if(temp->data == -1)
cout << temp->data << '\n';
}
else{
cout << temp->data << '\n';
temp = temp->prev;
PrintReverse(temp);
}
}
// returns the first element in the list and deletes the Node.
// caution, no error-checking here!
int popValue(){
node *n = head;
int ret = n->data;
head = head->next;
delete n;
return ret;
}
void swapN(node** a, node**b){
node*t = *a;
*a = *b;
*b = t;
}
node *head;
node *tail;
// Node *n;
};
/* A utility function to insert a node at the beginning of linked list */
void push(struct node** head_ref, int new_data)
{
/* allocate node */
struct node* new_node = (struct node*) malloc(sizeof(struct node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* A utility function to print linked list */
void printList(struct node *node)
{
while (node != NULL)
{
printf("%d", node->data);
node = node->next;
}
// printf("\n");
}
// A utility function to swap two pointers
void swapPointer( node** a, node** b )
{
node* t = *a;
*a = *b;
*b = t;
}
/* A utility function to get size of linked list */
int getSize(struct node *node)
{
int size = 0;
while (node != NULL)
{
node = node->next;
size++;
}
return size;
}
// Adds two linked lists of same size represented by head1 and head2 and returns
// head of the resultant linked list. Carry is propagated while returning from
// the recursion
node* addSameSize(node* head1, node* head2, int* carry)
{
// Since the function assumes linked lists are of same size,
// check any of the two head pointers
if (head1 == NULL)
return NULL;
int sum;
// Allocate memory for sum node of current two nodes
node* result = (node *)malloc(sizeof(node));
// Recursively add remaining nodes and get the carry
result->next = addSameSize(head1->next, head2->next, carry);
// add digits of current nodes and propagated carry
sum = head1->data + head2->data + *carry;
*carry = sum / 10;
sum = sum % 10;
// Assigne the sum to current node of resultant list
result->data = sum;
return result;
}
// This function is called after the smaller list is added to the bigger
// lists's sublist of same size. Once the right sublist is added, the carry
// must be added toe left side of larger list to get the final result.
void addCarryToRemaining(node* head1, node* cur, int* carry, node** result)
{
int sum;
// If diff. number of nodes are not traversed, add carry
if (head1 != cur)
{
addCarryToRemaining(head1->next, cur, carry, result);
sum = head1->data + *carry;
*carry = sum/10;
sum %= 10;
// add this node to the front of the result
push(result, sum);
}
}
// The main function that adds two linked lists represented by head1 and head2.
// The sum of two lists is stored in a list referred by result
void addList(node* head1, node* head2, node** result)
{
node *cur;
// first list is empty
if (head1 == NULL)
{
*result = head2;
return;
}
// second list is empty
else if (head2 == NULL)
{
*result = head1;
return;
}
int size1 = getSize(head1);
int size2 = getSize(head2) ;
int carry = 0;
// Add same size lists
if (size1 == size2)
*result = addSameSize(head1, head2, &carry);
else
{
int diff = abs(size1 - size2);
// First list should always be larger than second list.
// If not, swap pointers
if (size1 < size2)
swapPointer(&head1, &head2);
// move diff. number of nodes in first list
for (cur = head1; diff--; cur = cur->next);
// get addition of same size lists
*result = addSameSize(cur, head2, &carry);
// get addition of remaining first list and carry
addCarryToRemaining(head1, cur, &carry, result);
}
// if some carry is still there, add a new node to the fron of
// the result list. e.g. 999 and 87
if (carry)
push(result, carry);
}
node* reverse_list(node *m)
{
node *next = NULL;
node *p = m;
node *prev;
while (p != NULL) {
prev = p->prev;
p->prev = next;
next = p;
p = prev;
}
return prev;
}
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////
void Multiply2(node* n1, node* n2);
int digitsPerNode = 2;
node* result;
node* resultp = result;
node* resultp2 = result;
void Multiply(node* n1, node* n2)
{
if (n2->prev != NULL)
{
Multiply(n1, n2->prev);
}
Multiply2(n1, n2);
resultp2 = resultp = resultp->prev;
}
void Multiply2(node* n1, node* n2)
{
if (n1->prev != NULL)
{
Multiply2(n1->prev, n2);
}
if (resultp2 == NULL)
{
resultp2->data = 0;
result = resultp = resultp2;
}
int m = n1->data * n2->data + resultp2->data;
int carryon = (int)(m / pow(10, digitsPerNode));
resultp2->data = m % (int)pow(10, digitsPerNode);
if (carryon > 0)
{
if (resultp2->prev == NULL)
{
resultp2->prev->data = carryon;
}
else
{
resultp2->prev->data += carryon;
}
}
resultp2 = resultp2->prev;
}
/* int* buffer;
int lenBuffer = 0;
void multiplyHelper(int v, node* , int o);
void addToBuffer(int v, int i);
node* multiply(node* num1, node* num2)
{
if (num1 == NULL || num2 == NULL) return NULL;
int length1 = getSize(num1);
int length2 = getSize(num2);
if (length1 > length2) return multiply(num2, num1);
// initialize buffer
lenBuffer = length1 + length2;
buffer = new int[lenBuffer];
memset(buffer, 0, sizeof(int) * lenBuffer);
// multiply
int offset = 0;
node* anode = num1;
while (anode && anode->data!= -1)
{
multiplyHelper(anode->data, num2, offset);
anode = anode->prev;
offset++;
}
// transfer buffer to a linked list
node* h;
int pos = 0;
while (pos < lenBuffer && buffer[pos] == 0) pos++;
if (pos < lenBuffer)
{
node* temp;
temp->data = buffer[pos++];
h = temp;
anode = h;
while (pos < lenBuffer)
{
node* temp;
temp->data = buffer[pos++];
anode->prev = temp;
anode = anode->prev;
}
}
delete buffer;
lenBuffer = 0;
buffer = NULL;
cout << h->data << endl;
return h;
}
// multiply a single digit with a number
// called by multiply()
void multiplyHelper(int value, node* head, int offset)
{
// assert(value >= 0 && value <= 9 && head != NULL);
if (value == 0) return;
node* anode = head;
int pos = 0;
while (anode != NULL)
{
int temp = value * anode->data;
int ones = temp % 10;
if (ones != 0) addToBuffer(ones, offset + pos + 1);
int tens = temp / 10;
if (tens != 0) addToBuffer(tens, offset + pos);
anode = anode->prev;
cout << anode->data;
pos++;
}
}
// add a single digit to the buffer at place of index
// called by multiplyHelper()
void addToBuffer(int value, int index)
{
// assert(value >= 0 && value <= 9);
while (value > 0 && index >= 0)
{
int temp = buffer[index] + value;
buffer[index] = temp % 10;
value = temp / 10;
index--;
}
}*/
// Driver program to test above functions
int main(int argc, char *argv[])
{
char filename[50];
string name= argv[1];
string dig;
name.erase(0,9);//Parse input to only get input file.
ifstream file;
int digits;
for(int i = 0; i < name.length(); i++){
if(name.at(i) == ';'){
// dig = name.substr(0,name.length()-i);
name = name.substr(0,name.length()-i);
}
}
//cout << dig << endl;
//file.open("input.txt");
file.open(name.c_str());
digits = 2;
///////
///////////////////////////////////////////////////////////////////////
int words = 0;
int numbers = 0;
while(!file.eof()) //Goes through whole file until no more entries to input
{
string word;
getline(file,word); //Inputs next element as a string
// word << file;
//cout << word << '\n';
int x = 0;
node *head1 = NULL, *head2 = NULL, *result = NULL;
int counter = 0;
int t1index = 0; //keep tracks of nodes to multiply
int t2index = 0;
char operatorX;
LinkedList tempList1;
LinkedList tempList2;
while(x<word.length()) //Loops through each string input
{
//if(x<word.length()&&isalpha(word.at(x))) //Checks that x is in bounds and that char at position x is a letter
if(x<word.length()&&isdigit(word.at(x))) //Checks that x is in bounds and that char at position x is a number/digit
{
int start = x;
while(x<word.length()&&isdigit(word.at(x))) //Loops past the number portion
{
x++;
}
string temp = word.substr(start, x).c_str();
// cout << temp << '\n';
for(int i = 0; i < temp.length();i++){
tempList1.addValue(atoi(temp.substr(i, 1).c_str()));
// push(&head1, atoi(temp.substr(i, 1).c_str()));
counter++;
t1index++;
}
//search for the operator
while(x<word.length()){
if(x<word.length()&& (!isspace(word.at(x)) && !isdigit(word.at(x))))
{
while(x<word.length()&&(!isspace(word.at(x)) && !isdigit(word.at(x)))) //Loops past the letter portion
{
// cout << (word.at(x))<< '\n';
operatorX = word.at(x);
x++;
}
//search second value
while(x<word.length()){ //second value find
//start
if(x<word.length()&&isdigit(word.at(x))) //Checks that x is in bounds and that char at position x is a number/digit
{
int start = x;
while(x<word.length()&&isdigit(word.at(x))) //Loops past the number portion
{
x++;
}
string temp = word.substr(start, x).c_str();
for(int i = 0; i < temp.length();i++){
tempList2.addValue(atoi(temp.substr(i, 1).c_str()));
// push(&head2, atoi(temp.substr(i, 1).c_str()));
// cout << atoi(temp.substr(i, 1).c_str());
counter++;
}
//////START READING NUMBERS BACKWARDS
LinkedList finalList;
node* tempA = tempList1.tail;
node* tempB = tempList2.tail;
// multiply(tempA, tempB);
//ADDITION
while(tempA != NULL){
if(tempA->data != -1){
push(&head1,tempA->data);
// cout << tempA->data;
}
tempA = tempA->prev;
}
while(tempB != NULL){
if(tempB->data != -1){
push(&head2, tempB->data);
// cout << tempB->data;
}
tempB = tempB->prev;
}
// multiply(head1, head2);
// result = multiply(head1, head2);
// tempList1.PrintReverse();
addList(head1, head2, &result);
printList(head1);
cout << operatorX;
printList(head2);
cout << "=";
printList(result);
cout << endl;
}
else{
x++;
}
//end
}
}
else{
x++;
}
}
}
else //If char at position x is neither number or letter skip over it
{
x++;
}
}
}
}
最佳答案
由于您使用的是 C++,因此请使用模板/重载运算符。根据需要将 int
转换为浮点类型。参见例如: C++ Template problem adding two data types
关于c++ - 我有一个适用于整数的程序,但我需要让它也适用于小数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35692258/
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