Trying to figure out how can I make a program to identify equal numbers

Question

so my real question is, how can i make this code identify all the "look alike" numbers while theire running from 1 to 99, for example :11,22,33,44,... and while the program identify them it sends a message.

package doodle;

int num2=11;
for (int i=1; i<100; i++) {
    System.out.println(i);
    int num1=i;
    if(num1==num2) {
        System.out.println("WOW"); 
    }
}

Thanks

what exactly are you trying to do, and what exactly is(n't) working? why do you need this num1 variable? — Stultuske, Nov 05 '18 at 07:50
Possible duplicate of [Check if integer has repeating digits. No string methods or arrays](https://stackoverflow.com/questions/26748026/check-if-integer-has-repeating-digits-no-string-methods-or-arrays) — Mark, Nov 05 '18 at 07:51
If you mean numbers with repeating digits between 1 and 99, then that are only the numbers that are divisible by 11. — Johannes Kuhn, Nov 05 '18 at 07:56
What do you mean by "look alike"? numbers composed of twice the same digit? — Maurice Perry, Nov 05 '18 at 07:58

Scary Wombat · Accepted Answer · 2018-11-05T08:04:09.323

1

I would do using a String

for (int i = 11; i < 100; i++) {
    StringBuffer orig = new StringBuffer();
    String left = orig.append(i).toString();
    if (orig.reverse().toString().equals(left)) {
        System.out.println(left);
    }
}

or if you really wanted to use an int with flaky logic

int start = 11;
for (int i = 11; i < 100; i++) {
    if (i == start) {
        System.out.println(start);
        start += 11;
    }
}

Edit

As @mark has rightly pointed out, these solution only work whilst the range is up to 100

edited Nov 05 '18 at 08:04

answered Nov 05 '18 at 07:54

Scary Wombat

44,617
6
35
64

Note that the first method only works for `i < 100`, after `100` it will match with palindromes of numbers aswell like `121`, `131` etc.. Which is also what the question specified but it's good to know either way. – Mark Nov 05 '18 at 07:58
@Mark Yep, the requirements are not clear at all, but there maybe be a limit of 100 – Scary Wombat Nov 05 '18 at 08:00
@Mark While true, the domain here is `11 <= i < 100` (as the for loop states), so the problem doesn't exist – BackSlash Nov 05 '18 at 08:01
1

@BackSlash Just pointing it out if someone in the future comes across this answer and decides they want to use it for `i < 1000` or something – Mark Nov 05 '18 at 08:02
1

@Mark I have updated my answer with the disclaimer. – Scary Wombat Nov 05 '18 at 08:04

suvojit_007 · Answer 2 · 2018-11-05T08:00:07.310

1

int num2=11;
for (int i=1; i<100; i++) {
    if(i%num2==0) {  //<---- look alike
        System.out.println("WOW");
         }

edited Nov 05 '18 at 08:00

answered Nov 05 '18 at 07:54

suvojit_007

1,690
2
17
23

`121%11` would pass aswell. – Mark Nov 05 '18 at 07:55
check the range of i. i ranges from 1 to 99 – suvojit_007 Nov 05 '18 at 07:55
`if(i%num2==0)` – Scary Wombat Nov 05 '18 at 07:57
@ScaryWombat they are same – suvojit_007 Nov 05 '18 at 07:59
1

Yes, so why use a magic number in the code? BTW I did not downvote you. – Scary Wombat Nov 05 '18 at 07:59
@ScaryWombat thanks for pointing that out. I have fixed it. – suvojit_007 Nov 05 '18 at 08:00
Please also do not answer questions with code only answers. – Scary Wombat Nov 05 '18 at 23:49

score 0 · Answer 3 · edited Nov 05 '18 at 09:42

I would do it using String conversion and codePoint comparison

for (Integer number = 0; number < 1000; number++) {
    System.out.println(number);

    String stringnumber = String.valueOf(number);

    if (stringnumber.length() > 1 && stringnumber.codePoints().allMatch((digit) -> digit == stringnumber.codePointAt(0))) {
            System.out.println("WOW");
    }
}

length check (length() > 0) is needed to exclude all numbers with only one digit, otherwise, the program would print "WOW" for all numbers from 0 - 9 too.

All numbers from 0 to Integer.MAX_VALUE can be handled.

Trying to figure out how can I make a program to identify equal numbers

3 Answers3