How to determine if a List is sorted in Java?

Question

I would like a method that takes a List<T> where T implements Comparable and returns true or false depending on whether the list is sorted or not.

What is the best way to implement this in Java? It's obvious that generics and wildcards are meant to be able to handle such things easily, but I'm getting all tangled up.

It would also be nice to have an analogous method to check if the list is in reverse order.

Solution 1

Guava provides this functionality through its Comparators class.

boolean sorted = Comparators.isInOrder(list, comparator);

---

There's also the Ordering class, though this is mostly obsolete. An Ordering is a Comparator++. In this case, if you have a list of some type that implements Comparable, you could write:

boolean sorted = Ordering.natural().isOrdered(list);

This works for any Iterable, not just List, and you can handle nulls easily by specifying whether they should come before or after any other non-null elements:

Ordering.natural().nullsLast().isOrdered(list);

Also, since you mentioned that you'd like to be able to check for reverse order as well as normal, that would be done as:

Ordering.natural().reverse().isOrdered(list);

Solution 2

Stream

If you are using Java 8 or later, streams may be able to help.

list.stream().sorted().collect(Collectors.toList()).equals(list);

More briefly, in Java 16+, using Stream#toList.

list.stream().sorted().toList().equals(list);

This code will sort the list out of place and collect its elements in another list, which is then compared to the initial list. The comparison will be successful, if both lists contain the same elements in equal positions.

Note that this method will have worse space and time complexity than other approaches because it will have to sort the list out of place, so it should not be used for very large lists. But it is the easiest to use because it is a single expression and does not involve 3rd party libraries.

Solution 3

Here's a generic method that will do the trick:

public static <T extends Comparable<? super T>>
        boolean isSorted(Iterable<T> iterable) {
    Iterator<T> iter = iterable.iterator();
    if (!iter.hasNext()) {
        return true;
    }
    T t = iter.next();
    while (iter.hasNext()) {
        T t2 = iter.next();
        if (t.compareTo(t2) > 0) {
            return false;
        }
        t = t2;
    }
    return true;
}

Solution 4

Easy:

List tmp = new ArrayList(myList);
Collections.sort(tmp);
boolean sorted = tmp.equals(myList);

Or (if elements are comparable):

Object prev = null;
for( Object elem : myList ) {
    if( prev != null && prev.compareTo(elem) > 0 ) {
        return false;
    }
    prev = elem;
}
return true;

Or (if elements are not comparable):

Object prev = null;
for( Object elem : myList ) {
    if( prev != null && myComparator.compare(prev,elem) > 0 ) {
        return false;
    }
    prev = elem;
}
return true;

The implementations fail for lists containing null values. You have to add appropriate checks in this case.

Solution 5

If you need it for unit testing, you can use AssertJ. It contains an assertion to check if a List is sorted:

List<String> someStrings = ...
assertThat(someStrings).isSorted();

There is also an alternative method isSortedAccordingTo that takes a comparator in case you want to use a custom comparator for the sorting.

View more solutions

Comments

[Juliet]

Any chance you can use a sorted collection instead? Or just call .Sort() as needed instead?

[Eric Wilson]

The purpose of this is for testing, so no.

[João Portela]

is the list supposed to be in ascending or descending order?

[Eric Wilson]

I'd like to be able to check both, as both types will occur.

[Admin]

generics have nothing to do with sorting or comparability, they have to do with Type safety and that is all.

[Eric Wilson]

@fuzzy lollipop -- I was hoping to implement this method in type-safe way.

[Yishai]

Collections.sort would require that the list elements be comparable already, and that was a stipulation in the question in any event.

[BalusC]

This keeps continuing because previous will never be set. See answer of Daniel for right flow.

[Eric Wilson]

I think your first example is missing a line, since tmp is not populated.

[ColinD]

Would need to be constrained to <T extends Comparable> as well of course.

[mikera]

p.s. note the implementation is deliberately designed to avoid allocating an iterator or making multiple calls to get. This is designed for the common case where you are using an ArrayList, for other list implementation you are probably better off using the iterator.

[jjnguy]

@Colin (and everyone else) I am sorry that this doesn't compile. I wrote it very quickly. Colin, you have edit power, feel free to fix this error I made.

[ColinD]

The first example is also pretty wasteful for what it does, and the Objects in the second need to be Comparable.

[mikera]

Interestingly, your algorithm is actually O(1) expected time for reasonably random lists.

[ColinD]

One thing that could be done in this case would be to check if the List implements RandomAccess and use this implementation if so and an Iterator implementation if not. You wouldn't really want to use this with a LinkedList as is.

[Jesper]

It would be O(n) in the worst case (when the list is indeed sorted).

[mikera]

@ColinD great idea! I'm guessing you could do this so that it detects this at compile as well

[Kevin Bourrillion]

And there is also isStrictlyOrdered() if you want to make sure there are no duplicates.

[Kevin Bourrillion]

-1 since the asker has indicated he's getting "tangled up" in implementation issues like generics and wildcards, I think this answer is only telling him things he already knows.

[Eric Wilson]

I've accepted the solution that avoids using raw types. This was my second choice, though, so I've upvoted it.

[jjnguy]

@FarmBoy, Thanks for the feed back! Glad we could help.

[JUST MY correct OPINION]

+1 for pointing to existing code so that people stop reinventing things and start making new things.

[Rana Ghosh]

Can you show an example of this if there is no natural ordering and you want to pass in a Comparator? I think that's what Ordering.from is for: "Returns an ordering based on an existing comparator instance. Note that it is unnecessary to create a new anonymous inner class implementing Comparator just to pass it in here. Instead, simply subclass Ordering and implement its compare method directly."

[ColinD]

@tieTYT: You can using Ordering.from(comparator).isOrdered(list); if you're implementing the comparator yourself, though, you might as well just extend Ordering rather than implementing Comparator to begin with, in which case it's just myOrdering.isOrdered(list).

[Rann Lifshitz]

Following the suggestion to use Ordering.from(comparator).isOrdered(list): Using a Java 8 lambda expression in order to implement the comparator parameter of the from method can be a useful and elegant solution for customized ordering.

[Ben Page]

fwiw, the implementation of the library in the accepted answer (Guava) is essentially this: github.com/google/guava/blob/… But IMO it's always better to keep your codebase small rather than reimplementing things a library can give you.

[Eric Wilson]

Nice ... obviously didn't help 2010 me using Java 6, but this is a good addition.

[ShellFish]

This is so much better than sorting the array and checking if they are equal.

[Vadzim]

Much less effective than ideal allocation-less and sorting-less solution.

[johnlemon]

According to Java 8 docs, Collectors.toList(); There are no guarantees on the type, mutability, serializability, or thread-safety of the {@code List} returned; You should use a supplier of ArrayList

[Palle]

It does not look like this is relevant for this case, as the generated list never outlives is statement, so it will never be serialized, mutated or accessed from multiple threads.

[Moses Kirathe]

@Daniel Does prev need to be initialized?

[Andrea Bergonzo]

Just leaving a note here. Most of ordering is Obsolete but they removed the comments for planned deletion here. github.com/google/guava/commit/…

[Mingjiang Shi]

It doesn't work. ``` public static void main(String[] args) { List<Integer> list = Arrays.asList(1,2,9,7,4); System.out.println(isSorted(list)); } ``` expecte print false, but print true.

[Daniel]

@MosesKirathe: Yes