Java sort a list of pairs using both indexes [duplicate]

Question

I want to sort a List<Pair<Long, Long>> list = new ArrayList<Pair<Long, Long>>(); according to both the indexes, firstly according to first index and if first indexes are equal then according to second index using Java 8 lambda function.

I could easily sort only by first index as: Collections.sort(list,(o1, o2) -> o1.first < o2.first ? -1:0);

If I want to sort according to both the indexes Collections.sort(list,(o1, o2) -> o1.first < o2.first ? -1 : o1.first == o2.first ? (o1.second < o2.second ? -1 : 0) : 0); But I don't think its the right way to do it. Can someone please provide a better syntax?

Pair definition:

    class Pair<Type1, Type2> {
        Type1 first;
        Type2 second;

        Pair(Type1 f, Type2 s) {
            this.first = f;
            this.second = s;
        }

        public String toString() {
            return "(" + this.first + ", " + this.second + ")";
        }
    }

You can tell that a comparator is broken when you see that it sometimes returns a negative number but never a positive number. That obviously breaks the symmetry requirement. — Holger
– Holger, Commented Jun 3, 2021 at 7:20

Ravindra Ranwala · Accepted Answer · 2021-06-03 07:44:48Z

1

Use Collections.sort along with a comparator construction function like this:

Collections.sort(list, 
    Comparator.comparing((Pair<Long, Long> p) -> p.first)
        .thenComparing(p -> p.second));

Update

Alternatively, as suggested in the comment below you could use List.sort and it is a bit more succinct than the utility method used above.

list.sort(Comparator.comparing((Pair<Long, Long> p) -> p.first)
    .thenComparing(p -> p.second));

edited Jun 3, 2021 at 7:44

answered Jun 3, 2021 at 6:58

Ravindra Ranwala

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4 Comments

Holger Over a year ago

You can also use list.sort(…) directly.

Ravindra Ranwala Over a year ago

Of course, good catch. Sometimes it's confusing when the library authors provide different approaches of doing the same thing. However the suggested approach is much more succinct and worth mentioning as an update. Would the implementation specific to List use some specific optimizations over the utility method Collections.sort?

Holger Over a year ago

In case of the commonly used ArrayList, the sort method has been overridden to skip the copying step of the default method, however, since Java 8, update 20 the Collections.sort(…) method just delegates to list.sort(…), so you get the advantages of specialized implementations automatically. So, both will do the same, but list.sort(…) is just simpler.

Ravindra Ranwala Over a year ago

Yes, list.sort is just more readable I guess. Thanks for the detailed explanation.

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