How to sort the arry value in ascending order

Question

I have a string array where i need to sort the below values in ascending order in the same format...can anyone give me clue on this?
9.3
3.1
9.1
19.1
19
9.4
9.1.1
the sorted order should be
3.1
9.1
9.1.1
9.3
9.4
19
19.1

Answer 1

You may have easier luck writing your own comparator for this. These are headings in a table of contents, right?
Write a comparator that tokenizes a string on the '.' character, or use a StringBuffer to remove them, and then order the elements according to the combined numbers. Since alphanumeric would order this list as you want it anyway you could just order that way once the '.' are removed.
In other words your comparator would do this in the "compare" method:
public class MyComparator implements Comparator, java.io.Serializable {
private static Comparator stringComparator = java.text.Collator.getInstance();
...constructor(s), your own private "instance" variable of type MyComparator, a getInstance() method of your own, yadda yadda...
public int compare(Object item1, Object item2) {
try {
value1 = removePeriods((String)item1);
value2 = removePeriods((String)item2);
if (value1 == null) {
return (value2 == null) ? 0 : (-1);
return compare(value1, value2);
} catch (ClassCastException cce) {
System.err.println("Wrong Object Type, JackAss!");
protected int compare(String str1, String str2) {
return MyComparator.stringComparator.compare(str1, str2);
private String removePeriods(String value) {
StringBuffer sb = new StringBuffer(value);
int decimalIndex = value.indexOf('.');
while (decimalIndex != -1) {
sb.delete(decimalIndex, (decimalIndex + 1));
}