Read a grid by columns
A matrix is stored row by row, but a transpose report reads it column by column.
- Read columns from left to right, and within a column top to bottom.
- A ragged row contributes only the columns it actually has.
columnMajor(grid: list<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> columnMajor(List<List<Integer>> grid) {
}Worked examples
| Call | Result |
|---|---|
columnMajor(Main.<List<Integer>>ls(Main.<Integer>ls(1, 2), Main.<Integer>ls(3, 4))) | Main.<Integer>ls(1, 3, 2, 4) |
columnMajor(Main.<List<Integer>>ls(Main.<Integer>ls(1), Main.<Integer>ls(2, 3))) | Main.<Integer>ls(1, 2, 3) |
columnMajor(Main.<List<Integer>>ls(Main.<Integer>ls(1, 2, 3), Main.<Integer>ls(4, 5, 6))) | Main.<Integer>ls(1, 4, 2, 5, 3, 6) |
columnMajor(Main.<List<Integer>>ls(Main.<Integer>ls(5), Main.<Integer>ls(6), Main.<Integer>ls(7))) | Main.<Integer>ls(5, 6, 7) |
Hint
For each column position, walk every row and take the value when that row is long enough.
Reference solution in Java
List<Integer> columnMajor(List<List<Integer>> grid) {
List<Integer> result = new ArrayList<>();
int cols = 0;
for (List<Integer> row : grid) if (row.size() > cols) cols = row.size();
for (int c = 0; c < cols; c++) {
for (List<Integer> row : grid) if (c < row.size()) result.add(row.get(c));
}
return result;
}