Regex is the best tool for the job; what it should be depends on the problem specification. The following removes leading zeroes, but leaves one if necessary (i.e. it wouldn't just turn "0" to a blank string).

s.replaceFirst("^0+(?!$)", "")

The ^ anchor will make sure that the 0+ being matched is at the beginning of the input. The (?!$) negative lookahead ensures that not the entire string will be matched.

Test harness:

String[] in = {
    "01234",         // "[1234]"
    "0001234a",      // "[1234a]"
    "101234",        // "[101234]"
    "000002829839",  // "[2829839]"
    "0",             // "[0]"
    "0000000",       // "[0]"
    "0000009",       // "[9]"
    "000000z",       // "[z]"
    "000000.z",      // "[.z]"
};
for (String s : in) {
    System.out.println("[" + s.replaceFirst("^0+(?!$)", "") + "]");
}

See also

• regular-expressions.info
  • repetitions, lookarounds, and anchors
• String.replaceFirst(String regex)

If you really want a locally formatted date, you can use Date.parse and Date.format:

String shipDate = '04/06/2022';
System.debug(Date.parse(shipDate).format());

However, be aware that Date.parse depends on the user locale, and the format method also depends on the user locale. In other words, this answer will only work if the date is formatted in the correct locale for the user.

A more literal interpretation, we can split, format, and join, like this:

String shipDate = '04/06/2022';
String[] values = shipDate.split('/');
for(Integer i = 0; i < values.size(); i++) {
    values[i] = Integer.valueOf(values[i])+'';
}
System.debug(String.join(values,'/'));

For a Regular Expression approach, you can use:

String shipDate = '04/06/2022';
// Option 1:
shipDate = shipDate.replaceAll('\\b0(\\d)','$1');
// Option 2:
shipDate = shipDate.replaceAll('\\b0','');
System.debug(shipDate);

Note that if a Regular Expression wants an escaped character, such as \b or \d, you need to escape it again, because that's also Apex's escape character. This is why you see \\b and \\d in the example above.

Also note that Apex's Regular Expression format is closer to Java, not JavaScript, so /.../g doesn't actually do a global search, but just tries to match those characters. To enable the "g" flag, you have to do something like '(?g)\\b0' instead. Regardless, you don't need to, because replaceAll implies g.

You'll want to read Java's Pattern documentation for more information.

I am aware that we can use String Utils library. But what if the String has only "0". It returns " ". Do we have any other way to remove leading 0 for all other strings except for "0".

You can create your own utility method that does exactly what you want.

Well you still haven't answered my question about whether the department can be alphanumeric or just numeric.

Based on your examples you could just convert the String to an Integer. The toString() implementation of an Integer removes leading zeroes:

    System.out.println( new Integer("008").toString() );
    System.out.println( new Integer("000").toString() );
    System.out.println( new Integer("111").toString() );

If the string can contain alphanumeric the logic would be more complex, which is why it is important to define the input completely.

For an alphanumeric string you could do something like:

StringBuilder sb = new StringBuilder("0000");

while (sb.charAt(0) == '0' && sb.length() > 1)
    sb.deleteCharAt(0);

System.out.println(sb);

Or, an even more efficient implementation would be something like:

int i = 0;

while (product.charAt(i) == '0' && i < product.length() - 1)
    i++;

System.out.println( product.substring(i) );

The above two solutions are the better choice since they will work for numeric and alphanumeric strings.

Or you could even use the StringUtils class to do what you want:

String result = StringUtils.removeleadingZeroes(...) // whatever the method is

if (result.equals(" "))
    result = "0";

return result;

In all solutions you would create a method that you pass parameter to and then return the String result.