在Java中使用堆栈进行平衡的符号检查

Question

Hey guys so I have an assignment where Im supposed to read a string that the user inputs and check for balanced symbols using a stack. So if the string is "{[()]}" the string is balanced because there is a close for every opening. My thought is to use a loop that checks every character from the string that is given and if the string has an opener such as "([{" then it does stack.push(char) and if the character is a closer ")]}" then I need to use stack.pop(char). The issue im running into is that my professor is forcing me to use a string method and any help ive found online is using a boolean method, id appreciate it if someone could help me out here.

I understand that my code is not working but you can at least get the idea of what my logic is.

import java.util.*;

public class BalancedSymbols {

public static String balancedSymbols(String lineToCheck){ //this is the method that im being forced to use

    Stack<String> stack = new Stack<String>();

   for (int i = 0; i<lineToCheck.length(); i++){

        char x = '(';
        char y = '{';
        char z = '[';

        char a;
        a = lineToCheck.charAt(i);

        if (a == x){

            stack.push(a);

        }

        if (a == y){

            stack.push(a);

        }

        if (a == z){

            stack.push(a);

        }


    }

}

}

Obviously I would do the same for popping except with different symbols.

Answer 1

Here is the logic:

Create empty stack. (you already have done it).

Traverse through the String.

For every character ch you encounter, if ch is } ] ) and stack is empty return false
else

if it is push it in stack
if ch is check stack top .
if stack top is equal to corresponding opening bracket pop from stack else return false .
If you have reached to end of String and stack is not empty return false
else return true .
The reason behind returning false is we don't have corresponding matching parenthesis for this. These are extra brackets .

Try to implement it on your own first. If you face any problem, comment it.. I'll be happy to help.

Please comment if you have any questions.

Answer 2

 Create empty stack
 ==========================

private static boolean isValideEx(String str) {

    Stack<Character> st=new Stack<Character>();

    if(str == null || str.length() == 0)
        return true;

    for (int i = 0; i < str.length(); i++) {
        if(str.charAt(i)==')'){
                if(!st.isEmpty() && st.peek()=='('){
                    st.pop();
                }else{
                    return false;
                }
        }else if(str.charAt(i)==']'){
                if(!st.isEmpty() && st.peek()=='['){
                    st.pop();
                }else{
                    return false;
                }
        }else if(str.charAt(i)=='}'){
                if(!st.isEmpty() && st.peek()=='{'){
                    st.pop();
                }else{
                    return false;
                }
        }else{
            st.push(str.charAt(i));
        }

    }
    System.out.println(" sy "+st);

    if(st.isEmpty())
        return true;
    else
        return false;
}