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c++ - 将两个链表中的特定值相加时出现问题

转载 作者:太空宇宙 更新时间:2023-11-04 14:18:04 25 4
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你好,我有一个家庭作业,我需要将两个矩阵 .txt 文件读入两个链表(存储列、行和值),然后将两个列表相加并打印出求和矩阵。

将矩阵读入链表和将这些列表打印为矩阵一样工作正常。但是我坚持如何将这两个列表添加在一起。理想情况下,如果在将一个列表与另一个列表进行比较时列值和行值相同,则应将该值相加。如果它们不相同,那么它应该只打印值。我认为创建一个新列表的组合列表的大小然后比较元素并添加所有其他元素是可行的方法但我似乎无法让它比节点更先进。

在此先感谢您的帮助。

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <fstream>
#include <sstream>

using namespace std;

struct Node {
int row;
int column;
int value;
Node *next;
};
Node *A, *B, *C;

//Add node function.
void addNode( Node *& listpointer, int r, int c, int v ) {
Node *temp;
temp = new Node;
temp->row = r;
temp->column = c;
temp->value = v;
temp->next = listpointer;
listpointer = temp;
}

//Matrix row file size function.
int getRowSize ( char *file_name ){
int row = 0;
int nothing = 0;
ifstream input;
string line;
input.open(file_name);
if(!input.good()){
cout << "Cannot open file " << file_name << endl;
exit(0);
}
if(input.good()){
getline(input,line);
stringstream sline(line);
sline >> row >> nothing;
}
return row;
}

//Matrix column file size function.
int getColumnSize ( char *file_name ){
int column = 0;
int nothing = 0;
ifstream input;
string line;
input.open(file_name);
if(!input.good()){
cout << "Cannot open file " << file_name << endl;
exit(0);
}
if(input.good()){
getline(input,line);
stringstream sline(line);
sline >> nothing >> column;
}
return column;
}

//Read from file to LL function.
void readMatrix( Node* &a, char *file_name ){
int row = getRowSize(file_name);
int col = getColumnSize(file_name);
//cout << "Row = " << row <<" Column = "<< col <<endl;
int value = 0;
string line;
ifstream input;
input.open(file_name);
if(!input.good()){
cout << "Cannot open file " << file_name << endl;
exit(0);
}
if(input.good()){
getline(input,line);
stringstream sline(line);
sline >> row >> col;
//cout << "Matrix dimensions " << row << " " << col << endl;
}
for(int i = 0; i < row; ++i){
if(input.good()) {
getline(input,line);
stringstream sline(line);
for(int j = 0; j < col; ++j){
sline >> value;
if(value == 0) continue;
addNode(a, i, j, value);
//cout << "Element at (" << i << " " << j << ") is different than zero and it is: "<< value <<" \n";
}
//cout << endl;
}
}
input.close();
}

//Search function for print function.
int searchByPosition ( Node *listpointer, int r, int c){
Node *current;
current = listpointer;
while ( true ){
if ( current == NULL ){ break; }
if (r == current->row && c == current->column) {
//cout << "Value = " << x << "\n";
return current->value;
}
current = current->next;
}
//cout << "Value not in list.\n";
return 0;
}

//Print function.
void printMatrix ( Node *listpointer, int columnSize, int rowSize ){
int c, r, v;
for (r=0; r < rowSize; ++r) {
for (c=0; c < columnSize; ++c) {
v = searchByPosition(listpointer,r,c);
printf("%d ", v);
}
printf("\n");
}
printf("\n");
}

//Function that mneasures both lists and creates a new combined list.
void concatenate ( Node *&result, Node *listpointer1, Node *listpointer2){
Node *tempLP1, *tempLP2, *countLP1, *countLP2;
countLP1 = listpointer1;
countLP2 = listpointer2;
tempLP1 = listpointer1;
tempLP2 = listpointer2;
int listpointer1Size = 0;
int listpointer2Size = 0;
while ( countLP1 != NULL ) {
++listpointer1Size;
countLP1 = countLP1->next;
}
//cout << listpointer1Size <<endl;
while ( countLP2 != NULL ) {
++listpointer2Size;
countLP2 = countLP2->next;
}
//cout << listpointer2Size;
int resultSize = listpointer1Size + listpointer2Size;
//cout << resultSize;
for (int i=0; i < resultSize; ++i) {
if (tempLP1->column == tempLP2->column && tempLP1->row == tempLP2->row) {
//cout << result->value<<" "<<result->row<<" "<<result->column<<endl;
addNode(result, tempLP1->row, tempLP1->column, (tempLP1->value + tempLP2->value));
//cout <<"marker "<<i+1<<endl;
//result = result->next;
tempLP1 = tempLP1->next;
//cout << tempLP1->value<<" "<<tempLP1->row<<" "<<tempLP1->column<<endl;
//cout << result->value<<" "<<result->row<<" "<<result->column<<endl;
} else {
addNode(result, tempLP1->row, tempLP1->column, tempLP1->value);
//cout << tempLP1->value<<" "<<tempLP1->row<<" "<<tempLP1->column<<endl;
tempLP1 = tempLP1->next;
//addNode(result, listpointer2->row, listpointer2->column, listpointer2->value);
//cout <<"marker2"<<endl;
//cout << listpointer1->value;
//result = result->next;
//listpointer1 = listpointer1->next;
}
}

/*current = listpointer1;
prev = NULL;
while ( current != NULL ){
prev = current;
current = current->next;
}
if ( prev == NULL ){
//cout <<"List1 was empty, joining anyway.\n";
listpointer1 = listpointer2;
} else {
//cout <<"Join.\n";
prev->next = listpointer2;
}*/

}

int main() {
A = NULL; // ALL linked-lists start empty
B = NULL; // ALL linked-lists start empty
C = NULL; // ALL linked-lists start empty
readMatrix(A, (char*)"matrix1.txt");
readMatrix(B, (char*)"matrix2.txt");
int rowSize1 = getRowSize((char*)"matrix1.txt");
int colSize1 = getColumnSize((char*)"matrix1.txt");
//cout << rowSize << colSize;
printMatrix(A, rowSize1, colSize1);
printMatrix(B, rowSize1, colSize1);
concatenate(C, A, B);
//printMatrix(C, rowSize1, colSize1);
//printMatrix(B, rowSize1, colSize1);
}

最佳答案

下面是一些代码,可以将存储在链表中的两个 2 x 2 矩阵相加。

#include <stdio.h>
#include <stdlib.h>

typedef struct node {
int row;
int column;
int value;
struct node *next;
} Node;

void main()
{
int prev_node_row = 0;
Node * matrix1;
Node * matrix2;
Node * m1cur;
Node * m2cur;

Node m1n0 = {0,0,1,NULL};
Node m1n1 = {0,1,2,NULL};
Node m1n2 = {1,0,3,NULL};
Node m1n3 = {1,1,4,NULL};

Node m2n0 = {0,0,5,NULL};
Node m2n1 = {0,1,6,NULL};
Node m2n2 = {1,0,7,NULL};
Node m2n3 = {1,1,8,NULL};

matrix1 = &m1n0;
m1n0.next = &m1n1;
m1n1.next = &m1n2;
m1n2.next = &m1n3;

matrix2 = &m2n0;
m2n0.next = &m2n1;
m2n1.next = &m2n2;
m2n2.next = &m2n3;

m1cur = matrix1;
m2cur = matrix2;
while (m1cur != NULL && m2cur != NULL)
{
if (prev_node_row < m1cur->row)
{
printf("\n");
}
printf("%d ", m1cur->value + m2cur->value);
prev_node_row = m1cur->row;
m1cur = m1cur->next;
m2cur = m2cur->next;
}
}

初始化矩阵 1

1 2
3 4

初始化矩阵 2

5 6
7 8

输出

6 8
10 12

关于c++ - 将两个链表中的特定值相加时出现问题,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/9693986/

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