Sorting a String list with Integer values in Java

Question

I have a column VALUE in my table that contains:

`M_SYSCONFIG = 200600,2600000,700000600,110000600,150000600`

When I sort this list the result is: 110000600,150000600,110000600,200600,2600000,700000600

However, I need the list to be sorted as follows (treat the strings as integers): 200600,2600000,110000600,150000600,700000600

This is the code I have right now for sorting the list:

    JSONArray jsArray = dbcon.callSelectRecords("SELECT CODE, VALUE FROM M_SYSCONFIG WHERE MODULE = 'LIMIT_CONFIG' AND CODE in (?,?,?,?) ORDER BY VALUE", ft_other_cn, ft_own_account, payment, purchase);

    for (int i = 0; i< jsArray.size(); i++) {
        JSONObject js = JSON.newJSONObject(jsArray.get(i).toString());
        String trasactionType = JSON.get(js, "CODE");
        String value = JSON.get(js, "VALUE");
        List<String> data = Arrays.asList(value.split(","));
        Collections.sort(data);

I need to obtain the results as strings because after sorting I want to apply the following code:

        StringBuilder sbValue = new StringBuilder();
        if(ft_other_cn.equals(trasactionType)) {
            long limitOtherCimb = limit.getFtOtherCimbLimit();
            sbValue.append(limitOtherCimb).append(",");
            for(String values:data) {
                Long limitSysConfig = null;
                try {
                    limitSysConfig = Long.parseLong(values);
                } catch (Exception e) {}
                if(limitSysConfig == null) {
                    continue;
                }
                if(limitSysConfig > limitOtherCimb) {
                    continue;
                }
                sbValue.append(limitSysConfig).append(",");
            }
            customerLimit.setFtOtherCnLimit(StringUtils.removeEnd(sbValue.toString(), ","));

Instead of having list of strings create list of numerical types (possibly Long or even BigInteger depending on how big your numbers are). Then sorting it should be easier. — Pshemo
– Pshemo, Commented May 23, 2018 at 7:09
Of-topic : You read the DB and directly get a JSON, then work on the JSON to create a sorted list ? Why the JSON ? — AxelH
– AxelH, Commented May 23, 2018 at 7:17

Muddassir Rahman · Accepted Answer · 2018-05-23 07:15:21Z

3

You need to convert you string values to integers like this and then need to sort.

JSONObject js = JSON.newJSONObject(jsArray.get(i).toString());
    String trasactionType = JSON.get(js, "CODE");
    String value = JSON.get(js, "VALUE");
    List<String> data = Arrays.asList(value.split(","));
    List<Integer> intList = new ArrayList<Integer>();
    for(String s : data){
        intList.add(Integer.valueOf(s));
    }
    Collections.sort(intList);

answered May 23, 2018 at 7:15

Muddassir Rahman

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4 Comments

Yeru Over a year ago

i need String result, because i want to concat after that sorting.

Muddassir Rahman Over a year ago

@YeruAdi after sorting you can again convert to string values

Yeru Over a year ago

can you help me dude ? i try to convert but i can't

Yeru Over a year ago

can you help me dude ? i try to convert but i can't @Muddassir Rahman

Sumit raj · Accepted Answer · 2018-05-23 07:57:50Z

I suggest using biginteger because your numbers seems quite large. It's not the most efficient and optimized solution but yeah it will work

public static List<String> sortData(List<String> data){
            List<BigInteger>convertedData=new ArrayList<BigInteger>();
        for (String s : data) 
        {
            //System.out.println(s);
         convertedData.add(new BigInteger(s));
        }
        Collections.sort(convertedData);
        List<String>sortedData=new ArrayList<String>();

        for (BigInteger b : convertedData) 
        {
            sortedData.add(String.valueOf(b));

        }
        return sortedData;
    }

Your code:

JSONArray jsArray = dbcon.callSelectRecords("SELECT CODE, VALUE FROM M_SYSCONFIG WHERE MODULE = 'LIMIT_CONFIG' AND CODE in (?,?,?,?) ORDER BY VALUE", ft_other_cn, ft_own_account, payment, purchase);

for (int i = 0; i< jsArray.size(); i++) {
    JSONObject js = JSON.newJSONObject(jsArray.get(i).toString());
    String trasactionType = JSON.get(js, "CODE");
    String value = JSON.get(js, "VALUE");
    List<String> data = Arrays.asList(value.split(","));
    List<String> sortedData=sortData(data);  **<------**

Farvardin · Accepted Answer · 2018-05-23 07:16:45Z

1

Implement a Comparator like this:

Collections.sort(data, new Comparator<String>() {
            @Override
            public int compare(String o1, String o2) {
                return new Long(Long.parseLong(o1)).compareTo(Long.parseLong(o2));
            }
        });

answered May 23, 2018 at 7:16

Farvardin

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3 Comments

Scary Wombat Over a year ago

this sounds rather slow

Yeru Over a year ago

i am sorry, because i am new here dude :)

Amongalen Over a year ago

That looks to be quite suboptimal. Casting all elements to int/long, comparing them and then casting back to strings would most likely be faster. If I'm correct you're performing about log(n) times more casts.

Shubham Kadlag · Accepted Answer · 2018-05-23 08:29:13Z

1

You can take help of streams introduced in java 8.

Just add the below line after creating the List and you would have sorted string list

List<String> data = Arrays.asList(value.split(","));

data=data.stream().mapToLong(Long::parseLong).sorted().mapToObj(String::valueOf).collect(Collectors.toList());

If you very large numbers you can use BigInteger

data=data.stream().map(BigInteger :: new ).sorted().map(String::valueOf).collect(Collectors.toList());

If you are using java 6,7 you would have to use a comparator as mentioned by Taher

Collections.sort(data, new Comparator<String>() {
            @Override
            public int compare(String o1, String o2) {
                return new Long(Long.parseLong(o1)).compareTo(Long.parseLong(o2));
            }
        });

edited May 23, 2018 at 8:29

answered May 23, 2018 at 8:07

Shubham Kadlag

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Comments

wumpz · Accepted Answer · 2018-05-24 05:35:16Z

1

If you are not able to change your list, then the simplest way is to build a comparator using Java 8 and use the string values as bigintegers. You do not need to convert your string list to a number list.

List<String> list = Arrays.asList("200600,2600000,700000600,110000600,150000600".split(","));

list.sort(Comparator.comparing(item -> new BigInteger(item)));

System.out.println(list);

The magic happens within

Comparator.comparing(item -> new BigInteger(item))

With this you are constructing a Comparator (which is needed for sorting), that compares all items converted to BigIntegers.

answered May 24, 2018 at 5:35

wumpz

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Comments

Erwin Bolwidt · Accepted Answer · 2018-05-23 07:46:04Z

0

You are sorting the numbers as Strings - as a string, 11 comes before 2. You need to first convert the array of strings to numbers, then sort them as numbers.

With the Streams API you can do that on one line:

String value = ...;
List<Long> data = Arrays.stream(value.split(",")).map(Long::new).sorted()
                     .collect(Collectors.toList());

Since you need them as Long later, I'm using Long as the numeric type.

edited May 23, 2018 at 7:46

answered May 23, 2018 at 7:21

Erwin Bolwidt

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