Here is my code for the method, does not give me errors but is not working when calling it out, any help would be appreciated thanks:

static boolean isBalanced(String expr){
Stack stack = new Stack();

    for (char ch : expr.toCharArray()) {
        
        if (ch == '{' || ch == '(' || ch == '[') {
            stack.push(expr);

            
            } else {
                if (expr.isEmpty()) {
                                return false;
                        }
                
                    
                char latestOpenedPar = (char)stack.pop();
            
                if (latestOpenedPar == '{' && ch != '}') { 
                    return false;
                } else if (latestOpenedPar == '(' && ch != ')') { 
                    return false;
                } else if (latestOpenedPar == '[' && ch != ']') { 
                    return false;
                }
            }
                
            }

    return stack.isEmpty();
  

}

CodePudding user response：

If the string is "(abc)" you are going to push the ( and then pop it four times, once for each of a, b, c, and ), which of course will not work. Also expr.isEmpty() will never return true inside the loop because if it were empty, expr.toCharArray() would have returned a zero-length array and the loop would execute zero times. Use peek() to look at the top of the stack without removing it.

Also, your stack should be declared as:

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

in which case you would not need to cast the value returned by pop() or peek(); it would be automatically unboxed from a Character to a char.

CodePudding user response：

It is not very clear from the question but I guess you want to understand if the different brackets are "consistent".

1. if (expr.isEmpty()) {return false;} this is useless because if the condition were true it would never enter the for loop
2. if (ch == '{' || ch == '(' || ch == '[') {stack.push(expr);} here you put in the stack the entire string taken as input but after you comparing stack.pop() with the various brackets, so the right statement can be stack.push(ch);
3. You do stack.pop() without checking that the stack has at least one element