The challenge
Reverse Number is a number which is the same when reversed.
For example, the first 20 Reverse Numbers are:
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0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101
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Task
- You need to return the nth reverse number. (Assume that reverse numbers start from 0 as shown in the example.)
Notes
The solution in Java
Option 1:
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import java.math.BigInteger;
public class Palindrome {
private static int length = 0;
public static BigInteger findReverseNumber(long n) {
if (n == 1) {
return BigInteger.ZERO;
}
length = 0;
String remainder = findLengthAndRemainder(n - 1);
while (remainder.length() < (length + 1) / 2) {
remainder = "0" + remainder;
}
StringBuilder result = new StringBuilder();
if (remainder.matches("\\d0*")) {
result.append(remainder.charAt(0));
result.append("9".repeat(remainder.length() - 1));
} else {
result.append(Integer.parseInt(remainder.substring(0, 1)) + 1);
String value = String.valueOf(new BigInteger(remainder.substring(1)).subtract(BigInteger.ONE));
while (value.length() < remainder.length() - 1) {
value = "0" + value;
}
result.append(value);
}
String firstHalf = result.toString();
String secondHalf = result.reverse().substring(length % 2);
return new BigInteger(firstHalf + secondHalf);
}
private static String findLengthAndRemainder(long number) {
long subtract = 9;
while (number > 0) {
length++;
number -= subtract;
if (number <= 0) {
break;
}
length++;
number -= subtract;
if (number > 0) {
subtract *= 10;
}
}
number += subtract;
return String.valueOf(number);
}
}
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Test cases to validate our solution
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import org.junit.jupiter.api.Test;
import java.math.BigInteger;
import static org.junit.jupiter.api.Assertions.assertEquals;
class PalindromeTest {
@Test
void testFixed() {
assertEquals(new BigInteger("0"), Palindrome.findReverseNumber(1));
assertEquals(new BigInteger("1"), Palindrome.findReverseNumber(2));
assertEquals(new BigInteger("9"), Palindrome.findReverseNumber(10));
assertEquals(new BigInteger("909"), Palindrome.findReverseNumber(100));
assertEquals(new BigInteger("900000000000000000009"), Palindrome.findReverseNumber(100000000000L));
}
}
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