Rotate a list
A carousel shows the same items starting from a different one each time it advances.
- Rotating left by one moves the first item to the end.
- A shift larger than the list wraps around; a negative shift rotates the other way.
- An empty list rotates to an empty list.
rotateLeft(values: list<int>, by: int) → list<int>
Java needs a compiler and Drill does not host one yet, so this page is the reference rather than an exercise: the problem, worked examples, and the solution in full. To type it out, the same problem runs in Python.
Where you start
List<Integer> rotateLeft(List<Integer> values, int by) {
}Worked examples
| Call | Result |
|---|---|
rotateLeft(Main.<Integer>ls(1, 2, 3, 4), 1) | Main.<Integer>ls(2, 3, 4, 1) |
rotateLeft(Main.<Integer>ls(1, 2, 3, 4), 5) | Main.<Integer>ls(2, 3, 4, 1) |
rotateLeft(Main.<Integer>ls(1, 2, 3, 4), -1) | Main.<Integer>ls(4, 1, 2, 3) |
rotateLeft(Main.<Integer>ls(1, 2, 3), 0) | Main.<Integer>ls(1, 2, 3) |
Hint
Reduce the shift with a modulo first, and remember that the modulo of a negative number is negative in most of these languages.
Reference solution in Java
List<Integer> rotateLeft(List<Integer> values, int by) {
int n = values.size();
List<Integer> result = new ArrayList<>();
if (n == 0) return result;
int k = ((by % n) + n) % n;
for (int i = 0; i < n; i++) result.add(values.get((i + k) % n));
return result;
}