Rotate matrix elements for ArrayList<ArrayList<String>> in java

Question

I already did search a lot and got a lot of solutions for the rotate matrix but all of them are 2D arrays , for example arr[][] but for my case, my matrix is ArrayList<ArrayList<String>> . So it is natural that it might not be equal in row and column. For example, my matrix is -

1-2-3
4-5-6-6 //Here column is 4 not 3 like the 1-2-3 or 7-8-9 rows 
7-8-9

My target is to make my ArrayList<ArrayList<String>> matrix rotate into a clockwise or anti-clockwise and make it equal in a row and column size. For example-

1-2-3-0         1-4-7
4-5-6-6   ==>   2-5-8
7-8-9-0         3-6-9
                0-6-0

How to optimally achieve this purpose?

Answer 1

Assuming your original list elemnts are always different from 0 you could iterate over each column and collect the elemnts of each row and add 0 for any row which size is to short and stop when all elemnts equal 0 . Something like:

public static void main(String[] args) {

    List<List<String>> matrix = List.of( List.of("1", "2", "3"),
                                         List.of("4", "5", "6", "6"),
                                         List.of("7", "8", "9"));

    List<List<String>> rotated =
            IntStream.iterate(0, i -> i + 1)
                     .mapToObj(i -> matrix.stream()
                                          .map(sublist -> i < sublist.size() ? sublist.get(i) : "0")
                                          .toList())
                     .takeWhile(list -> !list.stream().allMatch(e -> "0".equals(e)))
                     .toList();
    
    rotated.forEach(System.out::println);
}

Answer 2

public List<List<String>> rotate(List<List<String>> matrix) {
    int maxRowLength = matrix.stream().map(List::size)
            .max(Comparator.naturalOrder()).orElse(0);
    return IntStream.range(0, maxRowLength)
            .mapToObj(i -> matrix.stream()
                    .map(l -> l.size() <= i ? "0" : l.get(i)).toList())
            .toList();
}

Explanation

This approach starts by getting the length of the longest row in the matrix:

matrix.stream().map(List::size).max(Comparator.naturalOrder()).orElse(0)

It then creates a stream which iterates through all possible list indexes: the ints between 0 and the max row length:

IntStream.range(0, maxRowLength)

For each index i in that list, it maps the index to a list. Each element of this list is the i th element of the corresponding list in the matrix, or "0" if that list is shorter than i :

.mapToObj(i -> matrix.stream()
        .map(l -> l.size() <= i ? "0" : l.get(i)).toList())

Finally, it converts the stream to a list:

.toList();

Answer 3

Here is a "Old Dog" approach:

public static List<List<Integer>> rotate (List<List<Integer>> source) {
   // find row with greatest number of columns
   // there will be one row generated for each column
   int max = 0;
   for (int i = 0; i < source.size(); ++i) {
      max = Math.max (source.get(i).size(), max);    
   }

   List<List<Integer>> dest = new ArrayList<> (max);
      // i indexes column in source, row in dest
     for (int i = 0; i < max; ++i) { 
        List<Integer> line = new ArrayList<> (source.size());
         // j indexes row in source, column in dest
        for (int j = 0; j < source.size(); ++j) { 
           if (i >= source.get(j).size()) {
               line.add (j,0);
           } else {
              line.add (j,source.get(j).get(i));
           }
        }
        dest.add (line);
   }     
   return dest;    
 }

Test method:

public static void testRotate () {
    Integer [][] arr = {{1,2,3},{4,5,6,10,11},{7,8,9}};
    List<List<Integer>> list = new ArrayList<> ();
    for (int i = 0; i < arr.length; ++i) {
        List<Integer> row = Arrays.asList (arr[i]);
        list.add(row);
    }
    List<List<Integer>> res = rotate (list);
    for (int i = 0; i < res.size(); ++i) {
        System.out.println ();
        for (int j = 0; j < res.get(i).size(); ++j) {
            System.out.print (res.get(i).get(j) + "\t");
        }
    }
    System.out.println ("\n");
}

An empty row ( {} ) in arr will result in a corresponding column in res having all zero.