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Java program for alphanumeric sorting objects

Java program for alphanumeric sort objects

please let me know how i can get the expected result

Expected Result: Bi 2 D1 D2 D14 E2

Actual output: Bi 2 D1 D14 D2 E2

====================================

List<Name> lst = new ArrayList<>();
lst.add(new Name("D",1));
lst.add(new Name("D",14));
lst.add(new Name("D",2));
lst.add(new Name("E",2));
lst.add(new Name("B",2));

Collections.sort(lst, new Comparator<Name>() {

    @Override
    public int compare(Name n1, Name n2) {
        // TODO Auto-generated method stub
        String o1=n1.getNm()+n1.getSeatnum();
        String o2=n2.getNm()+n2.getSeatnum();


         return o1.compareTo(o2);

    }
});

for (Name name : lst) {
    System.out.println(name.getNm()+name.getSeatnum());
}

=================================

public class Name {

    private String nm;
    private int seatnum;

    public int getSeatnum() {
        return seatnum;
    }

    public void setSeatnum(int seatnum) {
        this.seatnum = seatnum;
    }

    public Name(String nm) {
        super();
        this.nm = nm;
    }

    public Name(String nm, int seatnum) {
        super();
        this.nm = nm;
        this.seatnum = seatnum;
    }

    public String getNm() {
        return nm;
    }

    public void setNm(String nm) {
        this.nm = nm;
    }

}
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3 answers


Just compare letters and then integers:



public int compare(Name n1, Name n2) {
    // TODO Auto-generated method stub
    int compare = n1.getNm().compareTo(n2.getNm());
    if (compare == 0) {
        return Integer.compare(n1.getSeatnum(), n2.getSeatnum());
    } else {
        return compare;
    }
}
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Yes, it is possible, but your function compare

must first check the part String

for equality and then use a numeric comparison for the part of the number (currently both compare lexically ). So, you can use something like:



public int compare(Name n1, Name n2) {
    int c = n1.getNm().compareTo(n2.getNm());
    if (c != 0) {
        return c;
    }
    return Integer.valueOf(n1.getSeatnum()).compareTo(n2.getSeatnum());
}
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You can rewrite your comparator to work in 2 steps:

Collections.sort(lst, new Comparator<Name>() {
    @Override
    public int compare(Name n1, Name n2) {
        // compare the name part
        int nameCompare = n1.getName().compareTo(n2.getName());
        if(nameCompare != 0)
            return nameCompare;
        // compare the number part
        return n1.getSeatnum() - n2.getSeatnum();
    }
});

If you want to know the values null

, you must add:

Collections.sort(lst, new Comparator<Name>() {
    @Override
    public int compare(Name n1, Name n2) {
        // check for null Name
        if(n1 == null && n2 == null)
            return 0;
        else if(n1 == null)
            return -1;
        else if(n2 == null)
            return 1;

        // check for null in nx.getName()
        if(n1.getName() == null && n2.getName() == null)
            return 0;
        else if(n1.getName() == null)
            return -1;
        else if(n2.getName() == null)
            return 1;

        // compare the name part
        int nameCompare = n1.getName().compareTo(n2.getName());
        if(nameCompare != 0)
            return nameCompare;

        // compare the number part
        return n1.getSeatnum() - n2.getSeatnum();
    }
});

This approach would put the values ​​at the top of the list null

. If you prefer to have them at the end of the list, just replace 1

and -1

.

If you also want case insensitivity, change the name comparison string to:

        int nameCompare = n1.getName().toLowerCase().compareTo(n2.getName().toLowerCase());
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