所以我的作业要求我们使用双向链表来将数字相加或相乘并打印出来。我能够让它适用于整数，但是我无法弄清楚要改变什么以使其适用于十进制数。这是我到目前为止所得到的。我知道它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东西 例如，如果我执行50382 + 9281或482891 * 29734，此程序将正常工作，但我需要将其用于4.9171 + 49.2917或423.135 * 59
之类的工作编辑：假装int值是双倍的。我在我的实际代码上改了它，但是当我做数学时的结果仍然给了我一个整数，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++;
}
}
}
}
答案 0 :(得分：2)
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。见例如：
C++ Template problem adding two data types