Compare two strings

Question

I have the following code:

    int t = s.length()-1;
    int g = 0;

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

        if (s.charAt(i) != h.charAt(t--));
            g++;

    }

    if (g==0)
        return true;

    else 
        return false;

Basically, what this code should do is check if string h is inverse equal to string s, or vice versa. For some reason, “falsehood” always returns, although the obvious answer is correct.

Can anyone tell me what happened to my code?

Thank!

Answer 1

I would say that extra ;is the culprit.

Instead

if (s.charAt(i) != h.charAt(t--));

using

if (s.charAt(i) != h.charAt(t--))

You should always take a "safe" route. That is, use curly braces after if-else statements (and almost everywhere you can use them), so errors like this will not happen in the first place. The correct way to write:

if (s.charAt(i) != h.charAt(t--)) {
   g++;
}

, , , , s h .

Answer 2

extra ; if (s.charAt(i) != h.charAt(t--));

if (s.charAt(i) != h.charAt(t--))
{
  g++;
  break; // if not match, not need to continue with loop
}

Answer 3

If this is not a training exercise, I would advise you not to write the loops themselves and use some kind of library code. You can do:

String s = "abcd";
String h = "dcba";

System.out.println( h.equals( new StringBuffer(s).reverse().toString() ) );

or StringUtils # reverse .

---

Under the hood, these loops go through the string the same way you do. The code is in AbstractStringBuilder if you want to take a look.

Answer 4

use break;to exit loop if it doesn't have the same char

bool g = true;
for(.....)
{
   if (s.charAt(i) != h.charAt(t--))
   {
      g = false;
      break;
   }
}
return g;

It increases productivity.