how can i do this so better?!

[HearT.Hunt3r]

this is question :
Return a version of the given string, where for every star (*) in the string the star and the chars immediately to its left and right are gone. So "ab*cd" yields "ad" and "ab**cd" also yields "ad".

starOut("ab*cd") → "ad"
starOut("ab**cd") → "ad"
starOut("sm*eilly") → "silly"
===========
and this is my answer

public String starOut(String str) {
		StringBuilder string = new StringBuilder(str);
		String res = "";
		for(int i=0;i<string.length();i++){
			if(string.length()==1 && string.charAt(0)=='*')
	    		return "";
			if(i==0 && string.charAt(i)=='*'){
				string.deleteCharAt(i);
				string.deleteCharAt(i);
				i = 0;
			}else if(i==string.length()-1 && string.charAt(i)=='*'){
				string.deleteCharAt(i-1);
				string.deleteCharAt(i-1);
			}else if(i<=string.length() && string.charAt(i)=='*' && string.charAt(i+1)=='*' && string.charAt(i+2)=='*' ){
						string.deleteCharAt(i);
						string.deleteCharAt(i);
						string.deleteCharAt(i);
						string.deleteCharAt(i);
						string.deleteCharAt(i-1);
						i = i-2;
					    	}else if(i<=string.length() && string.charAt(i)=='*' && string.charAt(i+1)=='*' ){
								string.deleteCharAt(i);
								string.deleteCharAt(i);
								string.deleteCharAt(i);
	
								string.deleteCharAt(i-1);
								i = i-2;
							    	}else if(i<=string.length() && string.charAt(i)=='*'){
										string.deleteCharAt(i);
										string.deleteCharAt(i);
										string.deleteCharAt(i-1);
										i = i-2;
									    	}
			
		    }
		res += string;
		return res;
		}

(it's correct but it's very long and i think very idiot)!
how can i do this better?

[Junky]

ACK!

Never mind I cannot read.

[Junky]

The simplest way to achieve this would be to use indexOf and substring.

[HearT.Hunt3r]

And This is all exam:
Expected This Run
starOut("ab*cd") → "ad" "ad" OK
starOut("ab**cd") → "ad" "ad" OK
starOut("sm*eilly") → "silly" "silly" OK
starOut("sm*eil*ly") → "siy" "siy" OK
starOut("sm**eil*ly") → "siy" "siy" OK
starOut("sm***eil*ly") → "siy" "siy" OK
starOut("stringy*") → "string" "string" OK
starOut("*stringy") → "tringy" "tringy" OK
starOut("*str*in*gy") → "ty" "ty" OK
starOut("abc") → "abc" "abc" OK
starOut("a*bc") → "c" "c" OK
starOut("ab") → "ab" "ab" OK
starOut("a*b") → "" "" OK
starOut("a") → "a" "a" OK
starOut("a*") → "" "" OK
starOut("*a") → "" "" OK
starOut("*") → "" "" OK
starOut("") → "" "" OK
other test OK

[HearT.Hunt3r]

but how can i find all of the world by indexOf ?

[Junky]

What? Huh? Please explain.

[HearT.Hunt3r]

we have like this for ex : "vahidxyvahidxyvahidxy"
how can find all xy from this by indexOf?

[Junky]

How would you do it?

I mean with a pen and paper not using code.

[HearT.Hunt3r]

i donnu what did u said...anyway thx.

[Junky]

:sigh:

How would you do something more than once?

[JeffGrigg]

A couple of ideas:

1. Nested loops: When you find a '*', do another loop to find the end of the '*'s. Then delete characters.
2. Keep track of state: Store the index where you found the fist '*', and then continue going around the main loop until you find the last '*'. Then delete. But then you'll have to change the loop index, 'i', for it to come out right.
3. Copy characters from a String to a StringBuilder: Copy non-'*' characters as you find them (except in some cases). Probably have to keep track of the "previous character" found/processed. (If previousChar != '*' and thisChar == '*' then delete the last char from the output.)

[JeffGrigg]

And This is all exam:
Expected This Run
starOut("ab*cd") → "ad" "ad" OK
...
starOut("") → "" "" OK
other test OK

I like the automated tests. I like them a lot.

I also suggest adding this one:

starOut("ax*y*zd") → "ad" // "y" is deleted by both '*'s. (Be sure that the 2nd '*' does not delete the 'a'.)

[Junky]

A single while loop is all that is needed. Inside the loop you need an if statement to handle the four different situations: * is first char, * is last char, * is followed by another *, default of single * in the middle. Each condition will need to do 1 or 2 substring calls.

[Junky]

starOut("ax*y*zd") → "ad"

Grrrr!

My solution fails this. Time to rethink.

[Junky]

OK simpler solution is to only add a char to a StringBuilder if it is not followed or preceded by a *.

[HearT.Hunt3r]

OK simpler solution is to only add a char to a StringBuilder if it is not followed or preceded by a *.

Good Idea!

[JeffGrigg]

OK simpler solution is to only add a char to a StringBuilder if it is not followed or preceded by a *.

Hmmm...
Well, I tried it this way, and I tried it with my suggestion #3 above. And I refactored heavily, relying on automated unit tests. Much as I like my heavily computer science automata theory based state machine approach, I think I like the code that I got from Junky's suggestion better: It's more understandable, and hence more maintainable.

[HearT.Hunt3r]

i got that how i can find all word in a string by indexOf ..
int v=0;
for(int i=0;i<str.length();i++){
v = str.indexOf(word, i);
i += v;
}

[JeffGrigg]

The 'String.indexOf(char)' method can be used to find '*' character(s) in a string. So it could be used in this assignment.

However, the 'for' loop and 'charAt(i)' in your original posting will also work just as well. So I don't see a good reason for you to chase the 'indexOf' idea, unless you really want to.

[PavelChovanec]

The code below uses charAt() within a for-loop. It handled the 20 test strings I threw at it.

public String starOut(String str) {
     String result = "";
     boolean delete = false;
		  
     for (int i = 0; i < str.length(); i++) {
          if (str.charAt(i) == '*')
              delete = true;
          else if (delete)
              delete = false;
          else if (i == str.length() - 1 || str.charAt(i + 1) != '*')
              result = result + str.charAt(i);
     }
     return result;
}