Palindrome using stack

[Felicia123]

I've made my program become more advance due to my last thread using que & stack

#include<iostream>
#include<string>
#include<stack>
#include<queue>

using namespace std;

int main(){
	stack<char> s;
	queue<char> q;
	string str;
	int i = 0;
	char ch;

	cout << "Enter a text in lowercase: "; 
	getline(cin, str);
	cout << endl;

	for ( i = 0 ; i < str.length(); i++){
		ch = str[i];
		q.push(ch);
		s.push(ch);
	}

	while(!q.empty() && !s.empty()){
		if(q.front() != s.top()){
			cout << "The word " << str << " is not a palindrome" << endl;
			break;
		}
		else if( q.front() = s.top() ){
			i++;
			q.pop();
			s.pop();
			if( i = str.length()){
				cout << "The word " << str << " is a palindrome" << endl;
				break;
			}
		}
	}
	system( "pause" );
	return 0;
}

but my program only compare the last letter and first letter.
I read my slides but not like iterator . got begin and end . so it can compare using loop .
but how to do it through stack and queue?

[vlad from moscow]

In this statement

else if( q.front() = s.top() ){

s.top is assigned to q.front.:) But maybe it is a more advanced approach, who does know?

[vlad from moscow]

Also if the control will be passed inside tis code block

		else if( q.front() = s.top() ){
			i++;

i will become equal to str.length() + 1

[vlad from moscow]

As I said you early the more simple way though not so advanced as your is to use the following check

std::vector<char> v;

// some stuff to fill the vector

if ( v == std::vector<char>( v.rbegin(), v.rend() ) ) std::cout << "The word is polindrome\n";

[Felicia123]

do you mean?

else if( q.front() = str.length() + 1 ){
			i++;

is still fail to work..

[vlad from moscow]

I do not see any sense of using i in this loop. As for your new statement it is also invalid.

[Felicia123]

ya i know. but my question want me to do with queue and stack.

stack<char> s;
	queue<char> q;
	vector<char> v;
	int i = 0;
	char InputChar;


	//Enter sequence until user eof 
	do{
		//User input
		cout << "Enter string/sequence : ";
		cin  >> InputChar;

		//To let program won't repeat the same letter for the last chararacter
		//If without if , output : 123455
		//If statement  , output : 12345
		if( cin.good() )
			v.push_back(InputChar);
		else
			break;
	}while( !cin.eof() );

I'm using your way to fill up the vector


but now i've to use stack and queue to compare?
what's the algo?

[vlad from moscow]

What is the great sense to use a queue and a stack simultaneously? It seems that you should write two separate programs one that uses a queue and other that uses a stack.

[Felicia123]

but i see some people do using only 1 program.

http://answers.yahoo.com/question/index?qid=20091109124158AApRS6n


the link .
but i wan to know it. using my program..

[coder777]

It's ok to use the container to check the different behaviors.

Your problem is currently on line 30 and 34. You're using the assignment operator (... = ...) instead you certainly want to use the comparison operator (... == ...)

On line 30: You can omit if( q.front() = s.top() ) (not the else in front of it). Try to understand why it is so...

On line 31: You don't reset i to 0 after you used it in the loop on line 19

[Felicia123]

#include<iostream>
#include <vector>
#include<string>
#include<stack>
#include<queue>

using namespace std;

int main(){
	stack<char> s;
	queue<char> q;
	
	int i = 0;
	char InputChar;


	//Vector holds an object type char
	vector<char> v;
	istream::int_type c; 
  
	cout << "Enter a sequence of characters: "; 
	while ( ( c = cin.get() ) != EOF ) { 
		v.push_back( c ); 
	}

	if ( !v.empty() && v.back() == '\n' ) 
		v.pop_back(); 

	vector<char> v1(v.begin(),v.end());
	vector<char> v2(v.rbegin(),v.rend());

	for( i = 0 ; i < v.size() ; i++ ){
		s.push(v1[i]);
	}

	for( i = 0 ; i < v.size() ; i++ ){
		q.push(v2[i]);
	}

	if ( !v.empty() && v.back() == '\n' ) 
		v.pop_back(); 

	int i = 0; 
	while (  i < v.size() && q.front() == s.top() ) 
		i++; 
  
	cout << "The word "; 
  
	for ( char c : v ) 
		cout << c; 
	cout << " is" << ( i == v.size() ? "" :  " not" ) << " a polindrome." << endl;

here is the current code i done so far.

i having problem at my line 49 . as this line is teach by @vlad . any wrong at here?

[coder777]

line 49 is a language extension for C++11. You compiler likely doesn't support that as of yet.

You introduced a lot more errors now:

line 43: i is already defined (on line 13)
line 44: without pop() it always compares the very first element

line 30: imagine both q and s contain "abc". q.front() retrieves 'c'. s.top() retrieves 'a'. You used to use that to check whether it's a palindrome. But now you feed q with the reverse (cba) hence q.front() and s.top() both retrieve always the same

line 22: I have really no idea how EOF may appear within the cin stream

[Felicia123]

do you mean that i should change instead

vector<char> v2(v.rbegin(),v.rend());

to v2(v.begin() , b.end() )?

so what should i change for line 49?
so how many lines total i should change? getting blur a bit.

err . EOF that is no problem already . as my question stated it needed

[coder777]

Forget v1 and v2. Use v to feed s and q

so what should i change for line 49?

If your compiler doesn't support that replace it with an ordinary for:

  for ( i = 0; i < v.size(); i++) 
		cout << v[i]; 

within the while loop on line 44 you need to call pop() for both s and q

Line 43 doesn't give you a compiler error?

[Felicia123]

after i change the for() . only the error come out .
what mean use v to feed s and q?

isn't vector<char> v = q.front();?

something like that?

[Felicia123]

#include<iostream>
#include <vector>
#include<string>
#include<stack>
#include<queue>

using namespace std;

int main(){
	
	int i = 0;
	char InputChar;


	//Vector holds an object type char
	vector<char> v;
	istream::int_type c; 
  
	cout << "Enter a sequence of characters: "; 
	while ( ( c = cin.get() ) != EOF ) { 
		v.push_back( c ); 
	}

	if ( !v.empty() && v.back() == '\n' ) 
		v.pop_back(); 

	if ( !v.empty() && v.back() == '\n' ) 
		v.pop_back(); 

	stack<char> s;
	queue<char> q;

	while (  i < v.size() && q.front() == s.top() ){
		s.pop();
		q.pop();
		i++;
	}
  
	cout << "The word "; 
  
	for( i = 0 ; i < v.size() ; i++){
		cout << v[i];
	}

	cout << " is" << ( i == v.size() ? "" :  " not" ) << " a polindrome." << endl;

	system( "pause" );
	return 0;
}

having an error said that my
deque iterator are not dereferencable ?

[coder777]

with feed i mean providing data. You still need a loop to provide the data to s and q:

	for( i = 0 ; i < v.size() ; i++ ){
		s.push(v[i]);
		q.push(v[i]);
	}

Instead of this:

	if ( !v.empty() && v.back() == '\n' ) 
		v.pop_back(); 

	if ( !v.empty() && v.back() == '\n' ) 
		v.pop_back(); 

use a loop:

	while ( !v.empty() && v.back() == '\n' ) 
		v.pop_back();

Why don't you use your remove_if approach with isalnum:

http://www.cplusplus.com/reference/clibrary/cctype/isalnum/

[EDIT]

deque iterator are not dereferencable ?

runtime or compile time error?

[Felicia123]

it don't hv error when i push the s and q.
but now.

all my result are palindrome.
something wrong with my if else statement to compare the v[i]?

for my line 45?
why i get the result is a palindrome only?

[coder777]

The problem is that you use i in certain loops. when you check it on line 45 it is not longer what you think it is.

On line 33: use a for loop:

	for ( i = 0;  i < v.size() && q.front() == s.top(); i++ ){
		s.pop();
		q.pop();
	}
	bool is_pal = (i == v.size()); // here i is what you think. later not longer
...
	cout << " is" << ( is_pal ? "" :  " not" ) << " a polindrome." << endl; // do not use i here 

It'd be better if you use the index of a loop just internally (for(int i = 0; ...) and create variables for your decisions

[Felicia123]

i did it . after i just change to for loop . i get it.
because if
i++ . the original value will change also right?
so when go to the if statement . the value i is already changed ?