Sort ArrayList of Objects based on another array - Java

Question

I want to sort an ArrayList of Objects

nameArray = {[id:109,name:"abc"],[id:103,name:"bcd"],[id:105,name:"efg"],[id:102,name:"hij"],[id:111,name:"klm"]}

using another array

numberArray ={103,111}

now I want my sorted array to have values in the order

arrayAfterSort = {[id:103,name:"bcd"],[id:111,name:"klm"],... no matter other values in array can be of any order}

Can you please help me to do this using Java's Comparator.

• Turtle: What have you tried so far? Do you know how to use comparators?
• Sharan Rai: I know to use comparators when we need sort values of Array. i am not so clear about how to campare an array using another array.

Answer 1

A possible Java 8 solution:

nameArray.sort(Comparator.comparingInt(name -> {
 int index = numberArray.indexOf(name.id);
 return index == -1 ? Integer.MAX_VALUE : index;
 }));

Answer 2

This can be achieved using a custom comparator.

nameArray.sort(Comparator.comparing(MyClass::getId, numberArray::indexOf)).

Because indexOf returns -1 if it can't find a value those items will be first in the list. Your question said that their order doesn't matter but if you do want to force them to be last, or sort by some other criteria then, for example:

nameArray.sort(Comparator.comparing(MyClass::getId, numberArray::contains).reversed()
 .thenComparing(MyClass::getId, numberArray::indexOf)
 .thenComparing(MyClass::getName));

The first comparison returns a boolean which has a natural order of false first which needs to be reversed.

Answer 3

Try this.

List<Name> nameArray = Arrays.asList(
 new Name(109, "abc"),
 new Name(103, "abc"),
 new Name(105, "abc"),
 new Name(102, "abc"),
 new Name(111, "abc"));
List<Integer> numberArray = Arrays.asList(103, 111);
nameArray.sort(Comparator.comparing(n -> {
 int index = numberArray.indexOf(n.id);
 return index >= 0 ? index : Integer.MAX_VALUE;
}));
System.out.println(nameArray);

result:

[[id:103:name:abc], [id:111:name:abc], [id:109:name:abc], [id:105:name:abc], [id:102:name:abc]]