The challenge#

Convert an integer which is base 10, to a hexadecimal base 16 string.

Java provides various built-in options such as Integer.toString(a, 16), or String.format("0x%X", a), or Integer.toHexString(a)

The solution in Java code#

Option 1:

public static String convertToHex(int a){
return "0x"+Integer.toHexString(a).toUpperCase();
}
}

Option 2:

private static final String hexDigits = "0123456789ABCDEF";
public static String convertToHex(int a) {

while (a > 0) {
int digit = a % 16;
a = a / 16;
}
}
}

Option 3:

private static final String HEX = "0123456789ABCDEF";

public static String convertToHex(int a){
final StringBuilder sb = new StringBuilder("");
while (a != 0) {
sb.append(HEX.charAt((a & 0xf)));
a = a >> 4;
}
return "0x" + sb.reverse().toString();
}

}

Option 4:

private static final char[] numbers = {'0' , '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'};
public static String convertToHex(int a) {
if (a < 16)
return "0x" + numbers[a];
return convertToHex(a/16) + numbers[a%16];
}
}

Test cases to validate our solution#

import org.junit.Test;
import static org.junit.Assert.assertEquals;

@Test
public void test (){
}
}

import org.junit.Test;
import java.util.HashMap;
import static org.junit.Assert.assertEquals;

@Test
public void test (){

int numbers[] = new int[1000];
for (int i = 0; i < numbers.length; i++) {
numbers[i] = (int)Math.floor(Math.random()*4294967295.);
}

}

private class Hex {
public String convertToHex(int a) {
String num = "";
HashMap<String, String> mapeo = new HashMap<>();
for (int i = 0; i < 10; i++) mapeo.put(i + "", i + "");
mapeo.put("10", "A");
mapeo.put("11", "B");
mapeo.put("12", "C");
mapeo.put("13", "D");
mapeo.put("14", "E");
mapeo.put("15", "F");

while (a >= 16) {
num += mapeo.get(a % 16 + "") + "";
a /= 16;
}
num += mapeo.get(a + "");

return "0x"+(new StringBuilder(num)).reverse().toString();

}

}
}