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Comparator. How to sort by age, but if name is "A", move it to the end of list?

I have the next task:
I have object MyObject:
MyObject {
private String name;
private int age;
}
Using the Comparator compare() method implementation, I have to sort the list of MyObject by age, but if the name of MyObject is "A", this element should be moved to the end of the list. The list shouldn't contain 2 elements with "A" name, so we ignore this case.
I've written the next code, but it doesn't work
#Override
public int compare(MyObject myObject, MyObject t1) {
if (myObject.getName().equals("A")) {
return 1;
}else if (myObject.getAge() == t1.getAge()) {
return 0;
} else if (myObject.getAge() > t1.getAge()) {
return 1;
} else {
return -1;
}
}
Actually I don't surprised that it doesn't work, as I don't understand which sorting algorithm is used in this case.

Try something like:
#Override
public int compare(MyObject m1, MyObject m2) {
if (m1.getName().equals("A") && !m2.getName().equals("A")) {
return 1;
}else if (!m1.getName().equals("A") && m2.getName().equals("A")) {
return -1;
} else {
return Integer.compare(m1.getAge(), m2.getAge());
}
}

A Java Set for at most 2 objects

I'm writing a method that returns a Set<String>. The set may contain 0, 1, or 2 objects. The string keys are also quite small (maximum 8 characters). The set is then used in a tight loop with many iterations calling contains().
For 0 objects, I would return Collections.emptySet().
For 1 object, I would return Collections.singleton().
For 2 objects (the maximum possible number), a HashSet seems overkill. Isn't there a better structure? Maybe a TreeSet is slightly better? Unfortunately, I'm still using Java 7 :-( so can't use modern things like Set.of().
An array of 2 strings would probably give the best performance, but that's not a Set. I want the code to be self-documenting, so I really want to return a Set as that is the logical interface required.

Just wrap an array with an AbstractSet. You only have to implement 2 methods, assuming you want an unmodifiable set:
class SSet extends AbstractSet<String> {
private final String[] strings;
SSet(String[] strings) {
this.strings = strings;
}
#Override
public Iterator<String> iterator() {
return Arrays.asList(strings).iterator();
}
#Override
public int size() {
return strings.length;
}
}
If you want, you can store the Arrays.asList(strings) in the field instead of a String[]. You can also provide 0, 1 and 2-arg constructors if you want to constrain the array only to be that length.
You can also override contains:
public boolean contains(Object obj) {
for (int i = 0; i < strings.length; ++i) {
if (Objects.equals(obj, strings[i])) return true;
}
return false;
}
If you don't want to create a list simply to create an iterator, you can trivially implement one as an inner class:
class ArrayIterator implements Iterator<String> {
int index;
public String next() {
// Check if index is in bounds, throw if not.
return strings[index++];
}
public boolean hasNext() {
return index < strings.length;
}
// implement remove() too, throws UnsupportedException().
}

The set is then used in a tight loop with many iterations calling contains().
I would probably streamline it for this. Perhaps something like:
public static class TwoSet<T> extends AbstractSet<T> {
T a = null;
T b = null;
#Override
public boolean contains(Object o) {
return o.equals(a) || o.equals(b);
}
#Override
public boolean add(T t) {
if(contains(t)){
return false;
}
if ( a == null ) {
a = t;
} else if ( b == null ) {
b = t;
} else {
throw new RuntimeException("Cannot have more than two items in this set.");
}
return true;
}
#Override
public boolean remove(Object o) {
if(o.equals(a)) {
a = null;
return true;
}
if(o.equals(b)) {
b = null;
return true;
}
return false;
}
#Override
public int size() {
return (a == null ? 0 : 1) + (b == null ? 0 : 1);
}
#Override
public Iterator<T> iterator() {
List<T> list;
if (a == null && b == null) {
list = Collections.emptyList();
} else {
if (a == null) {
list = Arrays.asList(b);
} else if (b == null) {
list = Arrays.asList(a);
} else {
list = Arrays.asList(a, b);
}
}
return list.iterator();
}
}

You can achieve this by
Make a class that implements Set interface
Override add and remove method
Add value upon class initialisation by super.add(E element)
Use that class instead

Trying to compare the contents two Iterators, how?

EDIT: With your help I managed to fix my problem. I have edited my code to now show how I had to have it set up to get it working.
Currently I am having trouble coding a part which compares the content of two iterators. As part of the requirements for my assignment, I need to use a linkedlist to store the individual characters of the entered String. I have gotten to the point where I have two iterators which would contain the input one way and the reverse way.
String palindrom = input.getText();
String [] chara = palindrom.split (""); //this is successfully splitting them, tested.
int length = palindrom.length( ); // length == 8
System.out.println (length); //can use this for how many checks to do?
LinkedList ll = new LinkedList(Arrays.asList(chara));
Iterator iterator = ll.iterator();
Iterator desIterator = ll.descendingIterator();
/*while(iterator.hasNext() ){
System.out.println(iterator.next() );
}
while(desIterator.hasNext() ){
System.out.println(desIterator.next() );
}*/
boolean same = true;
while(iterator.hasNext()){
if(!iterator.next().equals(desIterator.next())){
same = false;
break;
}
}
And using the System.out I can see that they are being stored correctly, but I don't know how to check if the iterators store the same contents. What would be one of the simplest methods to compare the two iterators or convert them into something I can compare? To clarify I want to verify they contain the same elements in the same order.

boolean same = true;
while(iterator.hasNext()){
if(!desIterator.hasNext() || !iterator.next().equals(desIterator.next())){
same = false;
break;
}
}
System.out.println(same);

You need to iterate over both iterators simultaneously, i.e. with one loop. Here is a general comparison function (0 when equal, < 0 when A < B, > 0 when A > B):
static <T extends Comparable<S>, S> int compare(Iterator<T> a, Iterator<S> b) {
while (a.hasNext() && b.hasNext()) {
int comparison = a.next().compareTo(b.next());
if (comparison != 0) {
return comparison;
}
}
if (a.hasNext())
return 1;
if (b.hasNext())
return -1;
return 0;
}
To just check if they are equal, this can be simplified:
static <T, S> boolean equals(Iterator<T> a, Iterator<S> b) {
while (a.hasNext() && b.hasNext()) {
if (!a.next().equals(b.next())) {
return false;
}
}
if (a.hasNext() || b.hasNext()) {
// one of the iterators has more elements than the other
return false;
}
return true;
}
Guava implements this as Iterators.elementsEqual.

In both answers throw NullPointerException, if iterator.next() == null. This method is more optimal.
public static boolean equals(Iterator i1, Iterator i2) {
if (i1 == i2) {
return true;
}
while (i1.hasNext()) {
if (!i2.hasNext()) {
return false;
}
if (!Objects.equals(i1.next(), i2.next())) {
return false;
}
}
if (i2.hasNext()) {
return false;
}
return true;
}

Can Arrays. sort sort enumerations that have been added to a an array.?

I have an enum with the Ranks of a Deck. I add these Ranks into an array. When I used Arrays.sort to sort the array using a custom comparator it didn't sort instead it returned this:
Exception in thread "main" java.lang.NullPointerException at java.util.ComparableTimSort.countRunAndMakeAscending(ComparableTimSort.java:291)
at java.util.ComparableTimSort.sort(ComparableTimSort.java:171)
at java.util.ComparableTimSort.sort(ComparableTimSort.java:146)
at java.util.Arrays.sort(Arrays.java:472)
at HW3.Deck.sort(Deck.java:69)
at HW3.War.main(War.java:100)
Java Result: 1
Any suggestions? Thanks
Bellow is the code:
#Override
` public int compareTo(PlayingCard o) {
if (this==o) return 0;
if(this.getSuit()!=o.getSuit()){
return this.suit.compareTo(o.getSuit());
}
else {
return this.rank.compareTo(o.getRank());
}
}

If you use Arrays.sort(T[], Comparator), you'll have to take care of null values yourself.
Example
Arrays.sort(enums, new Comparator<Rank>>() {
#Override
public int compare(Rank o1, Rank o2) {
if (o1 == o2) {
return 0;
}
// Put NULL to the end.
if (o1 == null) {
return 1;
}
if (o2 == null) {
return -1;
}
return o1.compareTo(o2); // sort by ordinal
}
});

Sorted List in Java

I need to sort the list in java as below:
List contains collection of objects like this,
List list1 = {obj1, obj2,obj3,.....};
I need the final list which has "lowest value" and "repetition of name should avoid".
Ex:
List list1 = {[Nellai,10],[Gujarath,10],[Delhi,30],[Nellai,5],[Gujarath,15],[Delhi,20]}
After Sorting , I need the list like this :
List list1 = {[Nellai,5],[Gujarath,10],[Delhi,20]};
I have 2 Delhi (30,20) in my list. But I need only one Delhi which has lowest fare (20).
How to do that it in java?
Gnaniyar Zubair

If order doesn't matter, a solution is to use a Map[String, Integer], add an entry each time you find a new town, update the value each time the stored value is less than the stored one and then zip all the pairs into a list.

Almost the same as #Visage answer, but the order is different:
public class NameFare {
private String name;
private int fare;
public String getName() {
return name;
}
public int getFare() {
return fare;
}
#Override public void equals(Object o) {
if (o == this) {
return true;
} else if (o != null) {
if (getName() != null) {
return getName().equals(o.getName());
} else {
return o.getName() == null;
}
}
return false;
}
}
....
public Collection<NameFare> sortAndMerge(Collection<NameFare> toSort) {
ArrayList<NameFare> sorted = new ArrayList<NameFare>(toSort.size());
for (NameFare nf : toSort) {
int idx = sorted.getIndexOf(nf);
if (idx != -1) {
NameFare old = sorted.get(idx);
if (nf.getFare() < old.getFare()) {
sorted.remove(idx);
sorted.add(nf);
}
}
}
Collections.sort(sorted, new Comparator<NameFare>() {
public int compare(NameFare o1, NameFare o2) {
if (o1 == o2) {
return 0;
} else {
if (o1.getName() != null) {
return o1.getName().compareTo(o2.getName());
} else if (o2.getName() != null) {
return o2.getName().compareTo(o1.getName());
} else {
return 0;
}
}
}
});
}

I would do it in two stages.
Firstrly sort the list using a custom comparator.
Secondly, traverse the list and, for duplicate entries (which will now be adjacent to each other, provided you worte your comparator correctly), remove the entries with the higher values.

If you want to avoid duplicates, perhaps a class like TreeSet would be a better choice than List.

I would use an ArrayList like this:
ArrayList<Name> listOne = new ArrayList<Name>();
listOne.add(new Name("Nellai", 10);
listOne.add(new Name("Gujarath", 10);
listOne.add(new Name("Delhi", 30);
listOne.add(new Name("Nellai", 5);
listOne.add(new Name("Delhi", 20);
Collection.sort(listOne);
Then create the Name class
class name implements Comparable
{
private String name;
private int number;
public Name(String name, int number)
{
this.name= name;
this.number= number;
}
public String getName()
{
return this.name;
}
public int getNumber()
{
return this.number;
}
public int compareTo(Object otherName) // must be defined if we are implementing //Comparable interface
{
if(otherName instanceif Name)
{
throw new ClassCastException("Not valid Name object"):
}
Name tempName = (Name)otherName;
// eliminate the duplicates when you sort
if(this.getNumber() >tempName.getNumber())
{
return 1;
}else if (this.getNumber() < tempName.getNumber()){
return -1;
}else{
return 0;
}
}
}
I didn't compiled the code, it's edited here so you should fix the code. And also to figure out how to eliminate the duplicates and print only the lowest one.
You need to sweat too.