groovy simple way to find hole in list?

I'm using the grails findAllBy() method to return a list of Position(s). Position has an integer field called location, which ranges from 1 to 15. I need to find the lowest location in the position list that is free.

For example, if there are positions at locations 1开发者_C百科,2 and 4, then the algorithm should return 3. If locations 1 - 4 were filled, it would return 5.

Is there some simple groovy list/map functions to get the right number?

Thanks

If your list of positions were (limited to a mx of 5 for brevity):

def list = [ 1, 2, 4, 5 ]

And you know that you have a maximum of 5 of them, you can do:

(1..5).minus(list).min()

Which would give you 3

Just another option, because I originally thought he wanted to know the first unused slot in a list, say you had this:

def list = ['a', 'b', null, 'd', 'e', null, 'g']

You could easily find the first empty slot in the array by doing this:

def firstOpen = list.findIndexOf{ !it } // or it == null if you need to avoid groovy truth

Tim's way works, and is good for small ranges. If you've got the items sorted by location already, you can do it in O(n) by leveraging findResult

def firstMissing = 0
println list.findResult { (it.location != ++firstMissing) ? firstMissing : null }

prints 3.

If they're not sorted, you can either modify your db query to sort them, or add sort{it.location} in there.