Need Help in java. AGAIN! =)

[jairoh_]

---

using stacks and arrays
Given problem:
String Input: {this[is(just)a]test}
Elements inside the curly braces: [t, h, i, s, [, i, s, (, j, u, s, t, ), a, ], t, e, s, t]
Elements inside the brackets: [i, s, (, j, u, s, t, ), a]
Elements inside the parentheses: [j, u, s, t]

output of my program:
String Input: {this[is(just)a]test}
Elements inside the curly braces: [t, h, i, s, [, i, s, (, j, u, s, t, ), a, ], t, e, s, t]
Elements inside the brackets: [i, s, (, j, u, s, t, ), a, i, s, (, j, u, s, t, ), a, i, s, (, j, u, s, t, ), a, i, s, (, j, u, s, t, ), a, i, s, (, j, u, s, t, ), a, i, s, (, j, u, s, t, ), a]
Elements inside the parentheses: [j, u, s, t, j, u, s, t, j, u, s, t, j, u, s, t, j, u, s, t, j, u, s, t, j, u, s, t, j, u, s, t]

nag.balik2x ang output sa Elements inside the brackets ugElements inside the parentheses:
here is my source codes:

import java.util.*;
public class Brackets {
    private String input;
    private char [] array;
    private int index1=0,index2=0;
    private static Scanner in = new Scanner(System.in);
    private static Stack myStack = new Stack();
    public Brackets(){
        askInput();
        Process();
    }
    private void askInput(){
        System.out.println("String Input:");
        setInput(in.nextLine());
        setCharArray(new char [getInput().length()]);
    }
    private void Process(){
        for(int x =0; x<getInput().length();x++){
            char i = getInput().charAt(x);
            array[x] = i;
        }
        InsideTheCurlyBraces();
        InsideTheBrackets();
        InsideTheParentheses();
    }
    private void InsideTheCurlyBraces(){
        for(int x=0; x<getCharArray().length;x++){
            if(array[x]=='{'){
                index1 =x;
            }
            if(array[x]=='}'){
                index2 = x;
            }
            for(int y=index1+1; y<index2; y++){
            myStack.push(array[y]);
            }
        }
        
        System.out.println("Elements inside the curly braces: "+myStack);
        myStack.clear();
        index1 =0;
        index2 =0;
    }
    private void InsideTheBrackets(){
        for(int x=0; x<getCharArray().length;x++){
            if(array[x]=='['){
                index1 =x;
            }
            if(array[x]==']'){
                index2 = x;
            }
            for(int y=index1+1; y<index2; y++){
            myStack.push(array[y]);
            }
        }
        
        System.out.println("Elements inside the brackets: "+myStack);
        myStack.clear();
        index1 =0;
        index2 =0;
    }
    private void InsideTheParentheses(){
        for(int x=0; x<getCharArray().length;x++){
            if(array[x]=='('){
                index1 =x;
            }
            if(array[x]==')'){
                index2 = x;
            }
            for(int y=index1+1; y<index2; y++){
            myStack.push(array[y]);
            }
        }
        
        System.out.println("Elements inside the parentheses: "+myStack);
        myStack.clear();
        index1 =0;
        index2 =0;
    }
    private void setCharArray(char [] array){
        this.array = array;
    }
    private char [] getCharArray(){
        return array;
    }
    
    
    private void setInput(String input){
        this.input = input;
    }
    private String getInput(){
        return input;
    }
    
    public static void main(String args[]){
        Brackets b = new Brackets();
    }
    
}

anyone can help?tnx

[stealthghost]

question sa bro, {this[is(just)a]test}, mao ra na ang pattern?
1 curly, 1 bracket, 1 parenthesis at any order?

or pwede ni?

[{this{is[just](a){test}}(oh yeah)}]

[stealthghost]

ang

for(int y=index1+1; y<index2; y++){
     myStack.push(array[y]);
}

naka Nested Loop which is unnecessary since nangita ra ka kung asa ang character.
I-extract ang nested loop from the first loop like this.

for(int x=0; x<getCharArray().length;x++){
      if(array[x]=='('){
           index1 =x; //nangita ra ka sa first instance of (
      }
      if(array[x]==')'){
          index2 = x; //nangita ra ka sa last instance of (
     }
}
//gawas dapat kay mao man ang parameter nimo sa next loop.
//mag balik2x jud na imu code kay nag loop pa sa first
for(int y=index1+1; y<index2; y++){
     myStack.push(array[y]);
}

Btw, IMHO, it is better to reset at first, than sa last of the code like this.

private void InsideTheParentheses(){
        myStack.clear(); //reset here
        index1 =0; //reset here
        index2 =0; //reset here
        for(int x=0; x<getCharArray().length;x++){
            if(array[x]=='('){
                index1 =x;
            }
            if(array[x]==')'){
                index2 = x;
            }
        }
        for(int y=index1+1; y<index2; y++){
            myStack.push(array[y]);
        }
        System.out.println("Elements inside the parentheses: "+myStack);
    }

Since nag initialize ka balik sa value para makita nimo kung naunsa imung value. IMHO.

[jairoh_]

question sa bro, {this[is(just)a]test}, mao ra na ang pattern?
1 curly, 1 bracket, 1 parenthesis at any order?

or pwede ni?
bsag unsa bro. himo2x rato nako ang sample. kung unsai masud sa {}, [] ug () i output.

[{this{is[just](a){test}}(oh yeah)}]

ang

for(int y=index1+1; y<index2; y++){
     myStack.push(array[y]);
}

naka Nested Loop which is unnecessary since nangita ra ka kung asa ang character.
I-extract ang nested loop from the first loop like this.

for(int x=0; x<getCharArray().length;x++){
      if(array[x]=='('){
           index1 =x; //nangita ra ka sa first instance of (
      }
      if(array[x]==')'){
          index2 = x; //nangita ra ka sa last instance of (
     }
}
//gawas dapat kay mao man ang parameter nimo sa next loop.
//mag balik2x jud na imu code kay nag loop pa sa first
for(int y=index1+1; y<index2; y++){
     myStack.push(array[y]);
}

Btw, IMHO, it is better to reset at first, than sa last of the code like this.

private void InsideTheParentheses(){
        myStack.clear(); //reset here
        index1 =0; //reset here
        index2 =0; //reset here
        for(int x=0; x<getCharArray().length;x++){
            if(array[x]=='('){
                index1 =x;
            }
            if(array[x]==')'){
                index2 = x;
            }
        }
        for(int y=index1+1; y<index2; y++){
            myStack.push(array[y]);
        }
        System.out.println("Elements inside the parentheses: "+myStack);
    }

Since nag initialize ka balik sa value para makita nimo kung naunsa imung value. IMHO.

try nako ni bro. update lng ko ugma

[Klave]

Bai, I have to say, nagimprove gyud imong coding style.
Mas OOP na tanawon.

Otherwise kuwang ra nimo...
Usa nimo i-add check sa kung existing na ba.
Kung wala add, kung naa skip lang.

for(int y=index1+1; y<index2; y++){
    if(myStack.search(array[y]) == -1){
            myStack.push(array[y]);
    } }

Mao ni solution if you don't want to change your algorithm.
Otherwise, rather than looping the entire string again, you can just loop once and
You can have state or boolean variables to denote what is inside what.
Example, if you come across a open curly brace({), you turn insideCurlyBrace = true, then push all relevant characters inside a insideCurlyBraceStack, then turn insideCurlyBrace = false when you get the Close Curly Brace (}). Theoretically, a different state can be turned on in the middle of the loop and you push the character in different stacks at the same time. This also means you can have three different states set to true at the same time.

[jairoh_]

@stealthghost, ang pattern jud bro kai 1 curly, 1 bracket, 1 parenthesis in order like this = { [ ( ) ] },di man nuon ni assignment or project, exercise2x lng ba, mao nang nahan ko mu think outside the box kunohay.
@klave, kuha na nako bro, pero pasabta ko sa ani na code

for(int y=index1+1; y<index2; y++){
    if(myStack.search(array[y]) == -1){
            myStack.push(array[y]);
    } }

nganung == -1 man? hehehe tnx ninyu

[Klave]

matud pa sa Stack nga documentation.
Stack (Java 2 Platform SE v1.4.2)

Ang i-return sa search nga function kung dili niya makit-an kay -1.
Common ni siya basta mga search or lookup or matching nga function basta ang i-return kay ang index.
Kung makitan within the list, kung ikapila siya example kaduha siya, mao sad ireturn (= 2).
Otherwise, kung dili makit-an dili man sad pwede gamiton 0 kay sa ubang list or collection type nga object first object mana.
Mao -1 meaning walay nakit-an.

[jairoh_]

ah mao diay. tnx bro

[SuperStar]

habang tumatagal gumagaling ka na jairoh...

[johnmalkovich_cs1]

kani maoy thread!!