I need to sort a Set of String's which holds number.Ex: [15, 13, 14, 11, 12, 3, 2, 1, 10, 7, 6, 5, 4, 9, 8]. I need to sort it to [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]. But when i use Collections.sort(keyList); where keyList is Set, the reult i obtained is [1, 10, 11, 12, 13, 14, 15, 2, 3, 4, 5, 6, 7, 8, 9]. Please help.
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2String comparisons are alphabetic. Try to convert it to an Integer list to get the desired result.blackSmith– blackSmith2014-09-19 12:28:30 +00:00Commented Sep 19, 2014 at 12:28
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Please post the actual code that you used.wei2912– wei29122014-09-19 12:28:43 +00:00Commented Sep 19, 2014 at 12:28
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2Collections.sort accepts a Comparator as an argument. This enables you to define the compare function yourself. (Effectively, modify to int and then compare the values)Ronald– Ronald2014-09-19 12:30:51 +00:00Commented Sep 19, 2014 at 12:30
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6 Answers
Write a custom comparator and parse it as argument to Collections.sort(Collection,Comparator). One solution is parsing your Strings to Integers.
Collections.sort(keyList, new Comparator<String>()
{
@Override
public int compare(String s1, String s2)
{
Integer val1 = Integer.parseInt(s1);
Integer val2 = Integer.parseInt(s2);
return val1.compareTo(val2);
}
});
Comments
you could do as Kai said, and convert your String to integer and compare it
but it is expensive operation,what i suggest is this :
keyList.sort(new Comparator<String>() {
@Override
public int compare(String o1, String o2) {
if (o1.length() == o2.length()){
return o1.compareTo(o2);
}
return o1.length() - o2.length();
}
});
if your numbers has same length, then compare them by using String.compareTo, otherwise, sort them by order, so 1 2 3 will be automatically before 11 22 etc
Comments
can you try with:
final int[] searchList =
new int[] { 15, 13, 14, 11, 12, 3, 2, 1, 10, 7, 6, 5, 4, 9, 8 };
Arrays.sort(searchList);
The result is:
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
The list needs to be int
Comments
Your Strings will be sorted as Strings in natural order, and not as numbers. So, "11" comes after "10" and "2" will come after "11111111110".
What to do?.
Use Integer.parseInt() to parse each String value in the set as integer, then add them to a set and call Collections.sort().
Comments
Transform the Strings into Integers first.
List<Integer> ints = new ArrayList<>();
for (String s : strings)
ints.add(Integer.parseInt(s));
Collections.sort(ints);
If you don't require duplicate values, you can use a SortedSet, which maintains the order automatically:
SortedSet<Integer> ints = new TreeSet<>();
for (String s : strings)
ints.add(Integer.parseInt(s));
// all done!
Comments
Strings in Java are ordered lexicographically, comparing each character in order. Since '1' < '2', the string "12" is less than the string "2", since the first character of "12" is less than the first character of "2".
You can use a custom comparator to compare the numerical value of strings, rather than using their natural order:
Collections.sort(keyList, Comparator.comparing(Integer::parseInt));
