Groovy way to check if a list is sorted or not

Question

Does Groovy have a smart way to check if a list is sorted? Precondition is that Groovy actually knows how to sort the objects, e.g. a list of strings.

The way I do right now (with just some test values for this example) is to copy the list to a new list, then sort it and check that they are equal. Something like:

def possiblySorted = ["1", "2", "3"]
def sortedCopy = new ArrayList<>(possiblySorted)
sortedCopy.sort()

I use this in unit tests in several places so it would be nice with something like:

def possiblySorted = ["1", "2", "3"]
possiblySorted.isSorted()

Is there a good way like this to check if a list is sorted in Groovy, or which is the preffered way? I would almost expect Groovy to have something like this, since it is so smart with collections and iteration.

Answer 1

If you want to avoid doing an O(n*log(n)) operation to check if a list is sorted, you can iterate it just once and check if every item is less or equals than the next one:

def isSorted(list) {
    list.size() < 2 || (1..<list.size()).every { list[it - 1] <= list[it] }
}

assert  isSorted([])
assert  isSorted([1])
assert  isSorted([1, 2, 2, 3])
assert !isSorted([1, 2, 3, 2])

Answer 2

Why not just compare it to a sorted instance of the same list?

def possiblySorted = [ 4, 2, 1 ]

// Should fail
assert possiblySorted == possiblySorted.sort( false )

We pass false to the sort method, so it returns a new list rather than modifying the existing one

You could add a method like so:

List.metaClass.isSorted = { -> delegate == delegate.sort( false ) }

Then, you can do:

assert  [ 1, 2, 3 ].isSorted()
assert ![ 1, 3, 2 ].isSorted()