Nanyang Science and Technology OJ topic exercises---bracket pairing problem __ algorithm (data structure)

Here's my algorithm:

#include <iostream>
using namespace std;

int IsMatch (char* str)
{
	char stack[10000];
	int i=0;
	memset (stack,0,sizeof (stack));
	char *p=str;
	if (*p== '] ' | | *p== ') return
		0;
	while (*p!= ' ")
	{
		if (*p== '] ' | | | | *p== ') '
		) {
			switch (*p)
			{case
			'] ':
				if (* (p-1) = = ' [')
				{
					i--;
					stack[i]= ' n ';
				}
				else return 0;
				break;
			Case ') ':
				if (* (p-1) = = ' (')
				{
					i--;
					stack[i]= ' n ';
				}
				else return 0;
			}
		}
		else
		{
			stack[i++]=*p;
		}
		p++;
	}
	if (i==0 && stack[i]== ') return
		1;
	else return 
		0;

}

int main ()
{
	int N;
	Char str[10000];
	cin>>n;
	while (n--)
	{
       cin>>str;
	   if (IsMatch (str)) cout<< "Yes" <<endl;
	   else
	   cout<< "No" <<endl;
	}
	return 0;
}

Algorithm Reviews:

Personal feeling is to see your understanding of the stack, no difficulty

Attached OJ system based on memory footprint and time recommended by the optimal solution: (since I have not yet studied the STL, or back to study this algorithm)

#include <iostream>
#include <vector>
#include <string>
using namespace std;
int main ()
{
int n;
cin>>n;
while (n--)
{  
vector<char> vec;
string ch;
Vec.push_back (");
cin>>ch;
for (int i=0;i<ch.length (); i++)
{
vec.push_back (ch[i]);
if (Vec.back ()-1 = * (Vec.end ()-2) | | | Vec.back ()-2 = = (Vec.end ()-2))
{
vec.pop_back ();
Vec.pop_back ();
}
if (Vec.size () ==1)
cout<< "Yes" <<endl;
else
cout<< "No" <<endl;
}
return 0;
}