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balanced parenthesis function

1 reply on 1 page. Most recent reply: Apr 23, 2003 2:52 AM by Adam Duffy

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jake

Posts: 83
Nickname: onorok
Registered: May, 2002

balanced parenthesis function

Posted: Apr 22, 2003 10:29 PM

this is supposed to return true if there is a balanced amount of parenthesis like a valid mathmatical expression, I am using a stack.
see if you can find any problems cause it doesn't work!

public boolean hasBalParens(String s)           
	{						// this should return false:  "s)ss()a(dcf)a)"
		final char LEFT_PAREN='(';
		final char RIGHT_PAREN=')';
		StringBuffer str = new StringBuffer(s);
		
		if(str.charAt(0)!=RIGHT_PAREN)
		{	
			for(int i=0; i<str.length(); ++i)
			{
				if(str.charAt(i)==LEFT_PAREN)
				{
					parserStack.push("(");
				}
			
				if(str.charAt(i)==RIGHT_PAREN)
				{
					if(parserStack.peek()=="(")
						parserStack.pop();
					
				}
			}
		}
 
		if(parserStack.isEmpty() && str.charAt(0)!=RIGHT_PAREN)
		{
			return true;
		}
		else
			return false;
			
	}

Adam Duffy

Posts: 168
Nickname: adamduffy
Registered: Feb, 2003

Re: balanced parenthesis function

Posted: Apr 23, 2003 2:52 AM

Hi Jake,

I ran it with the example shown in the comments, i.e. s)ss()a(dcf)a) and it threw an exception at the

if(parserStack.peek()=="(")

line of code. The peek method of java.util.Stack will throw EmptyStackException if the Stack is empty. This will occur at the first right parenthesis - the second character in the string mentioned above.

Suggest you modify the line to read

if(!parserStack.isEmpty() && parserStack.peek()=="(")

which will make sure the stack is not empty before you peek into it.

Furthermore, you might consider returning false on the else condition of this if statement, i.e.

if(str.charAt(i)==RIGHT_PAREN)
{
    if(!parserStack.isEmpty() && 
        parserStack.peek()=="(")      
    {
        parserStack.pop();		
    }
    else
    {
        // at this point, we have a right parenthesis
        // without a left parenthesis coming before it
        return false;
    }
}

I do not see the need for the if statement outside the for loop, i.e.

if(str.charAt(0)!=RIGHT_PAREN)

You should be able to remove this without affecting your code.

All the best,
Adam

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