Interleave two lists
Two queues for two registers are interleaved so every other customer comes from each.
- Take one from the first list, then one from the second, and so on.
- When one list runs out, append whatever remains of the other.
alternatingMerge(first: list<int>, second: list<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> alternatingMerge(List<Integer> first, List<Integer> second) {
}Worked examples
| Call | Result |
|---|---|
alternatingMerge(Main.<Integer>ls(1, 2, 3), Main.<Integer>ls(9)) | Main.<Integer>ls(1, 9, 2, 3) |
alternatingMerge(Main.<Integer>ls(1), Main.<Integer>ls(4, 5)) | Main.<Integer>ls(1, 4, 5) |
alternatingMerge(Main.<Integer>ls(1, 2), Main.<Integer>ls(3, 4)) | Main.<Integer>ls(1, 3, 2, 4) |
alternatingMerge(Main.<Integer>ls(), Main.<Integer>ls(1, 2)) | Main.<Integer>ls(1, 2) |
Hint
Loop up to the longer length and take each element that still exists.
Reference solution in Java
List<Integer> alternatingMerge(List<Integer> first, List<Integer> second) {
List<Integer> result = new ArrayList<>();
int n = Math.max(first.size(), second.size());
for (int i = 0; i < n; i++) {
if (i < first.size()) result.add(first.get(i));
if (i < second.size()) result.add(second.get(i));
}
return result;
}