Answer to Question #228727 in C for Arup

Question

A matrix m x n that has relatively few non-zero entries is called sparse matrix. It may be represented in much less than m n space. An mxn matrix with k non-zero entries is sparse if k<<m <n. It may be faster to represent the matrix compactly as a list of the non-zero indexes and associated entries. WAP to represent a sparse matrix using linked list.

Answer 1

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <malloc.h>


typedef struct List
{
	int date;
	struct List* pNext;
}List;
void pushList(List** hd, const int d)
{
	if ((*hd) == NULL)
	{
		(*hd) = (List*)malloc(sizeof(List));
		(*hd)->pNext = NULL;
		(*hd)->date = d;
	}
	else
	{
		List* p = (*hd);
		while (p->pNext)
		{
			p = p->pNext;
		}
		List* nw = (List*)malloc(sizeof(List));
		nw->pNext = NULL;
		nw->date = d;
		p->pNext = nw;
	}
}
void PrintList(List* h,int col)
{
	List* p = h;
	int ind = 0;
	while (p)
	{
		printf("%i ", p->date);
		ind++;
		p = p->pNext;
	}
	//print zero
	for (int i = 0; i < col - ind; i++)
		printf("%i ", 0);
	printf("\n");
}
void clearList(List** h)
{
	List* pr = NULL;
	while ((*h)->pNext)
	{
		pr = (*h);
		(*h) = (*h)->pNext;
		free(pr);
	}
	free((*h));
}
int main()
{
	int m;
	int n;
	printf("Enter m,n:\n");
	scanf("%i%i", &m, &n);
	List** ls = (List*)malloc(sizeof(List*)*m);
	for (int i = 0; i < m; i++)
		ls[i] = NULL;
	printf("Enter elements:\n");
	for (int i = 0; i < m; i++)
	{
		for (int j = 0; j < n; j++)
		{
			int x;
			scanf("%i", &x);
			if (x != 0)
				pushList(&ls[i], x);
		}
	}
	printf("====================Output WAP List=====================\n");
	for (int i = 0; i < m; i++)
		PrintList(ls[i], n);
	for (int i = 0; i < m; i++)
		if(ls[i])
		clearList(&ls[i]);
	free(ls);
	return  0;
}