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

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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

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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

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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.