Java charAt() String index out of range: 5

Question

I am trying to figure out "what 5-digit number when multiplied by 4 gives you its reverse?" using this code but I get error: Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 5 at java.lang.String.charAt(String.java:658) at Digits.main(Digits.java:12)

 public class Digits{
  public static void main(String[] args) {
    int n = 0;
    int b = 0;
    String number = Integer.toString(n);
    String backwards = Integer.toString(b);

for (int x = 9999; x < 100000 ; x++ ) {
  n = x;
  b = x *4;

  if (number.charAt(0) == backwards.charAt(5 )&& number.charAt(1) == backwards.charAt(4)
  && number.charAt(2) == backwards.charAt(3) && number.charAt(3) == backwards.charAt(2)
  && number.charAt(4) == backwards.charAt(1) && number.charAt(5) == backwards.charAt(0)) {
    System.out.println(n);
    break;
  }
}

Any help would be grealy appreciated

Elliott Frisch · Accepted Answer · 2015-07-13T04:24:16.050

Correct. Because the first five characters are at indices 0, 1, 2, 3 and 4. I would use a StringBuilder (because of StringBuilder.reverse()). And, I would suggest you restrict variable visibility. Then remember to modify number and backwards when you change n and/or b. Something like

for (int x = 9999; x < 100000; x++) {
    int n = x;
    int b = x * 4;
    String number = Integer.toString(n);
    String backwards = Integer.toString(b);
    StringBuilder sb = new StringBuilder(number);
    sb.reverse();
    if (sb.toString().equals(backwards)) {
        System.out.printf("%s * 4 = %s", number, backwards);
    }
}

And I get

21978 * 4 = 87912

You're too fast me for me, Frisch! – Tim Biegeleisen Jul 13 '15 at 04:21 — Tim Biegeleisen, Jul 13 '15 at 04:21

score 0 · Answer 2 · answered Jul 13 '15 at 04:17

0

backwards and number are String, which internally uses an array. And an array are indexed from 0 to size-1 . Hence such statements will throw ArrayIndexOutOfBoundsException:

backwards.charAt(5 )
number.charAt(5)

answered Jul 13 '15 at 04:17

Juned Ahsan

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1

Your observation is correct, but this doesn't help with the problems in his code. – Tim Biegeleisen Jul 13 '15 at 04:21

score 0 · Answer 3 · answered Jul 13 '15 at 04:18

At the time you create your strings, both of your ints are 0, so both of your strings are "0" for the duration of your program. What you really want is the strings to change every time your number changes. So your code should look more like this:

public class Digits{
  public static void main(String[] args) {
    int n = 0;
    int b = 0;
    String number;
    String backwards;

for (int x = 10000; x < 100000 ; x++ ) {
  n = x;
  b = x *4;

  number = Integer.toString(n);
  backwards = Integer.toString(b)

  . . .
}

In addition, arrays in Java are zero-indexed, so for instance for the string "10000", your program will throw the index out of bounds exception on backwards.charAt(5) because the string is indexed from character 0 to character 4.

Java charAt() String index out of range: 5

3 Answers3

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