How do I sort a Set to a List in Java?

The answer provided by the OP is not the best. It is inefficient, as it creates a new List and an unnecessary new array. Also, it raises "unchecked" warnings because of the type safety issues around generic arrays.

Instead, use something like this:

public static
<T extends Comparable<? super T>> List<T> asSortedList(Collection<T> c) {
  List<T> list = new ArrayList<T>(c);
  java.util.Collections.sort(list);
  return list;
}

Here's a usage example:

Map<Integer, String> map = new HashMap<Integer, String>();
/* Add entries to the map. */
...
/* Now get a sorted list of the *values* in the map. */
Collection<String> unsorted = map.values();
List<String> sorted = Util.asSortedList(unsorted);

Sorted set:

return new TreeSet(setIWantSorted);

or:

return new ArrayList(new TreeSet(setIWantSorted));

List myList = new ArrayList(collection);
Collections.sort(myList);

… should do the trick however. Add flavour with Generics where applicable.

Always safe to use either Comparator or Comparable interface to provide sorting implementation (if the object is not a String or Wrapper classes for primitive data types) . As an example for a comparator implementation to sort employees based on name

    List<Employees> empList = new LinkedList<Employees>(EmpSet);

    class EmployeeComparator implements Comparator<Employee> {

            public int compare(Employee e1, Employee e2) {
                return e1.getName().compareTo(e2.getName());
            }

        }

   Collections.sort(empList , new EmployeeComparator ());

Comparator is useful when you need to have different sorting algorithm on same object (Say emp name, emp salary, etc). Single mode sorting can be implemented by using Comparable interface in to the required object.

There's no single method to do that. Use this:

@SuppressWarnings("unchecked")
public static <T extends Comparable> List<T> asSortedList(Collection<T> collection) {
  T[] array = collection.toArray(
    (T[])new Comparable[collection.size()]);
  Arrays.sort(array);
  return Arrays.asList(array);
}

@Jeremy Stein I wanted to implement same code. As well I wanted to sort the set to list, So instead of using Set I converted set values into List and sort that list by it's one the variable. This code helped me,

set.stream().sorted(Comparator.comparing(ModelClassName::sortingVariableName)).collect(Collectors.toList());

I am using this code, which I find more practical than the accepted answer above:

List<Thing> thingList = new ArrayList<>(thingSet);
thingList.sort((thing1, thing2) -> thing1.getName().compareToIgnoreCase(thing2.getName()));