如何用C语言将链表存储的稀疏矩阵以表格形式输出？

Got it, let's adjust your code so it outputs the matrix in a proper table format (including zeros) instead of just listing non-zero elements. Here's the step-by-step fix and modified code:

First, we need two key things:

• Find the maximum column index used in the matrix so we know how many columns to print
• For each row, iterate through every column from 0 to max column, printing the stored value if it exists, otherwise 0

Modified Full Code #

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

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

void Init(element *x[]) {
    int i;
    for (i = 0; i < 3; i++)
        x[i] = NULL;
}

// Helper function to find the maximum column index in the matrix
int getMaxColumn(element *x[]) {
    int maxCol = -1;
    for (int i = 0; i < 3; i++) {
        element *temp = x[i];
        while (temp != NULL) {
            if (temp->column > maxCol) {
                maxCol = temp->column;
            }
            temp = temp->next;
        }
    }
    // If matrix is empty, return 0 to avoid negative column count
    return maxCol == -1 ? 0 : maxCol;
}

void Insert(element *x[], int row, int column, int value) {
    int r = row;
    element *p;
    element *new = malloc(sizeof(element));
    new->row = row;
    new->column = column;
    new->value = value;
    if (x[r] == NULL) {
        x[r] = new;
        new->next = NULL;
    } else {
        p = x[r];
        if (new->column < p->column) {
            new->next = p;
            x[r] = new;
        } else if (new->column > p->column) {
            while (p->next != NULL && p->next->column < new->column) {
                p = p->next;
            }
            new->next = p->next;
            p->next = new;
        } else
            printf("An element already exists there!!\n");
    }
}

void Printout(element *x[]) {
    int maxCol = getMaxColumn(x);
    int isEmpty = 1;

    // Check if matrix is empty first
    for (int i = 0; i < 3; i++) {
        if (x[i] != NULL) {
            isEmpty = 0;
            break;
        }
    }
    if (isEmpty) {
        printf("This matrix is empty!!\n");
        return;
    }

    // Print each row as a table line
    for (int i = 0; i < 3; i++) {
        element *temp = x[i];
        for (int j = 0; j <= maxCol; j++) {
            // If current column has a value, print it and move to next node
            if (temp != NULL && temp->column == j) {
                printf("%d ", temp->value);
                temp = temp->next;
            } else {
                printf("0 "); // Print zero for empty columns
            }
        }
        printf("\n");
    }
}

int main(int argc, const char * argv[]) {
    int choice, column, row, value, number;
    element *a[3], *b[3], *sum[3];
    Init(a);
    Init(b);
    Init(sum);
    do {
        printf("\n***\tADDING SPARSE MATRICES\t***\n");
        printf("\n 1.) Insert in A");
        printf("\n 2.) Insert in B");
        printf("\n 3.) Printout both");
        printf("\n 0.) EXIT");
        printf("\nChoose ---------> ");
        scanf("%d", &choice);
        switch (choice) {
            case 1: /*Insert in A */
                do {
                    printf("Enter row -> ");
                    scanf("%d", &row);
                } while (row < 0 || row > 2); // Fixed: rows are 0-2 (since a[3] is 3 elements)
                do {
                    printf("Enter column -> ");
                    scanf("%d", &column);
                } while (column < 0);
                printf("Enter value -> ");
                scanf("%d", &value);
                Insert(a, row, column, value);
                break;
            case 2: /*Insert in B */
                do {
                    printf("Enter row -> ");
                    scanf("%d", &row);
                } while (row < 0 || row > 2);
                do {
                    printf("Enter column -> ");
                    scanf("%d", &column);
                } while (column < 0);
                printf("Enter value -> ");
                scanf("%d", &value);
                Insert(b, row, column, value);
                break;
            case 3: /* Printout A & B */
                printf("\n::::::: MATRIX A :> \n\n");
                Printout(a);
                printf("\n::::::: MATRIX B :> \n\n");
                Printout(b);
                break;
            default:
                printf("\nWRONG CHOICE");
        }
    } while (choice != 0);
    return 0;
}

Key Changes Explained #

1. getMaxColumn Function: This scans all nodes in the matrix to find the highest column index used. This tells us how many columns we need to print for each row.
2. Revised Printout Function:
   • First checks if the matrix is empty (same as your original logic)
   • For each row, loops through every column from 0 to the max column. For each column:
     • If there's a node in the current row with this column index, print its value and move to the next node
     • If not, print 0
3. Fixed Row Input Validation: In your original code, inserting into matrix A allowed rows up to 3, but your array only has indices 0-2 (since element *a[3] creates 3 rows). This was a potential crash bug, so I fixed the condition to row > 2.

Example Output #

If you insert these elements into Matrix A:

• Row 0, Column 1, Value 3
• Row 1, Column 1, Value 5
• Row 2, Column 0, Value 7; Row 2, Column 1, Value 8; Row 2, Column 2, Value 9

The output will look like:

::::::: MATRIX A :> 

0 3 0 
0 5 0 
7 8 9 

内容的提问来源于stack exchange，提问作者RogerSK