1Z0-829 Working with Arrays and Collections Practice Question
Which TWO of the following will sort a List<String> in natural (ascending) order?
⚠ Common exam trap
Many candidates confuse `Collections.sort(list)` (which sorts in natural order) with `Collections.sort(list, Collections.reverseOrder())` (which sorts in reverse order), and may also mistakenly think `Arrays.sort()` works on a `List` without realizing it requires an array.
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
list.sort(Comparator.naturalOrder());
`Comparator.naturalOrder()` returns a comparator that imposes the natural (ascending) ordering on `String` objects, which is lexicographic order based on Unicode values. The `List.sort()` method accepts this comparator and sorts the list in place, making it a concise and idiomatic way to sort a list in ascending order.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✓
list.sort(Comparator.naturalOrder());
Why this is correct
Correct: naturalOrder comparator sorts in ascending order.
- ✗
list.sort((a,b) -> b.compareTo(a));
Why it's wrong here
Incorrect: comparing b to a reverses the order, resulting in descending.
- ✗
Collections.sort(list, Collections.reverseOrder());
Why it's wrong here
Incorrect: reverseOrder sorts in descending order.
- ✓
Collections.sort(list);
Why this is correct
Correct: Collections.sort sorts the list in natural order.
- ✗
Arrays.sort(list);
Why it's wrong here
Incorrect: Arrays.sort does not accept a List; it works on arrays only.
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