In Java 8 you can do so:
strList.stream().map(File::new).collect(Collectors.toList())
Most complete example:
List<String> strList = new ArrayList<>();
//adiciona itens em strList
List<File> fileList = strList.stream().map(File::new).collect(Collectors.toList());
Explanation
First, strList.stream() generates a Stream from the list. Stream is the representation of a collection of elements that supports operations across the set. It is something like SQL or jQuery.
Then the method map of Stream allows you to apply an operation to the whole set. The operation we want to perform is to convert a String for a File. We do this by passing a method that converts a single element and the method map takes care to apply to all elements.
We could use a lambda method as follows:
List<File> fileList = strList.stream().map(s -> new File(s)).collect(Collectors.toList());
However, we can simplify this by passing the reference to the File who receives a String.
Note that we could pass any method that receives a String as parameter and return a File and get the same result.
Finally, we took the result of processing map, that transformed a Stream of String in a Stream of File and collected on a list of File usando o métodoCollect` and stating what kind of data structure we want.
Are you sure you want to File?
Since Java 7 it is recommended to use the new input and output API (NIO or New IO). Therefore, instead of using File you should be using Path.
Example:
List<Path> fileList = strList.stream().map(Paths::get).collect(Collectors.toList());