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>
C++ 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
std::vector<int> alternatingMerge(std::vector<int> first, std::vector<int> second) {
}Worked examples
| Call | Result |
|---|---|
alternatingMerge(std::vector<int>{1, 2, 3}, std::vector<int>{9}) | std::vector<int>{1, 9, 2, 3} |
alternatingMerge(std::vector<int>{1}, std::vector<int>{4, 5}) | std::vector<int>{1, 4, 5} |
alternatingMerge(std::vector<int>{1, 2}, std::vector<int>{3, 4}) | std::vector<int>{1, 3, 2, 4} |
alternatingMerge(std::vector<int>{}, std::vector<int>{1, 2}) | std::vector<int>{1, 2} |
Hint
Loop up to the longer length and take each element that still exists.
Reference solution in C++
std::vector<int> alternatingMerge(std::vector<int> first, std::vector<int> second) {
std::vector<int> result;
int n = (int) first.size();
if ((int) second.size() > n) n = (int) second.size();
for (int i = 0; i < n; i++) {
if (i < (int) first.size()) result.push_back(first[i]);
if (i < (int) second.size()) result.push_back(second[i]);
}
return result;
}