I am assuming you are talking about Java here.

So the loop will cost O(n) time, my question is that will Arrays.sort() cost more time?

**Yes**, `Arrays.sort(int[])`

in all Java standard library implementations that I know, is an example of a comparison-based sort and thus must have worst-case complexity Ω(n log n). In particular, Oracle Java 7 uses a dual-pivot quicksort variant for the integer overloads, which actually has an *Ω(n ^{2})* worst case.

and will Arrays.sort() cost more space?

In all likelihood it will use ω(1) space (which means another **yes**, the space usage is not O(1)). While it’s not impossible to implement a comparison-based sort with only constant extra space, it’s highly impractical.

That said, under certain conditions it is possible to sort specific types of data in linear time, see for example:

With a constant range of input integers (for example if `abs(A[i]) <= C`

for some constant C), then counting sort and radix sort use indeed only O(n) time and O(1) space, so that might be useful.