无法访问Java中的嵌套列表数组

Question

Sorry for the clunky title, English is not my first language.

I'm having issues controlling how a nested for loop is going around a list of lists.

Example: I have the letters, {A,B,C,D,E,F,G,H,I}. They are in the 2d list like this:

 List<List<Character>> alphabet = new ArrayList<List<Character>>(3);
 alphabet.get(0).add('A'); 
 alphabet.get(1).add('B'); 
 alphabet.get(2).add('C'); 
 alphabet.get(0).add('D'); 
 alphabet.get(1).add('E'); 
 alphabet.get(2).add('F'); 
 alphabet.get(0).add('G'); 
 alphabet.get(1).add('H'); 
 alphabet.get(2).add('I');

So every 3rd letter in the alphabet goes in the same list. I'm having a few problems making the for loop that would go through the alphabet list and reconstruct the alphabet in the correct order. Any help would be appreciated.

Which would be the best solution, to go across and get the letters in order, A then B then C then D.

Or go down each list first, get A then D then G, and when I move to the next list, slot then into the required position?

Thank you

// Rephrased

Imagine 3 lists, in a list. The 3 lists represent columns of letters.

alphabet = [[a, d, g, j], [b, e, h], [c, f, i]]

What is the best way to loop through the nested list, and reconstruct the alphabet in the correct order?

I will then put the result in a string and write it to a file.

This is my for loop.

for(int k = 0, i = 0; k < decryptedtextColumns.get(i).size(); k++)
{
    for(i = 0; i < decryptedtextColumns.size(); i++)
    {
        if(k <= decryptedtextColumns.get(i).size())
        {
            Character letter = decryptedtextColumns.get(i).get(k);
            decryptedtext.add(letter);
        }
    }
} 

Answer 1

Java 8 has a simple and short way to do it.

 List<Character> sortedAlphabets= alphabet.stream().flatMap(list->list.stream()).sorted().
                 collect(Collectors.toList());

produces output

[A, B, C, D, E, F, G, H, I]

Answer 2

You'll have to use the MOD operator and do arithmetics on char.

Here is a possible solution :

     for (int i = 0 ; i < 26 ; i++){
         alphabet.get(i%3).add((char) ('A' + i));
     }

• You actually use the MOD of i%3 to go systematically from 0 to 2 and add the value of char 'A' + i each time you loop.

Answer 3

You basically just need a multi-dimensional form of the merge function from the mergesort algorithm.

Check out the following code at IdeOne .

class Ideone
{
    public static void main (String[] args) throws java.lang.Exception
    {
        List<List<Character>> alphabet = new ArrayList<List<Character>>(3);
        List<Character> first = new ArrayList<Character>();
        first.add('a');
        first.add('d');
        first.add('g');
        first.add('i');
        List<Character> second = new ArrayList<Character>();
        second.add('b');
        second.add('e');
        second.add('h');
        List<Character> third = new ArrayList<Character>();
        third.add('c');
        third.add('f');
        third.add('j');
        alphabet.add(first);
        alphabet.add(second);
        alphabet.add(third);

        List<Character> mergedSoFar = new ArrayList<Character>();
        for(int i = 0; i < alphabet.size(); ++i) {
            List<Character> sortedLetters = alphabet.get(i);
            mergedSoFar = merge(mergedSoFar, sortedLetters);
        }
        System.out.println(mergedSoFar);
    }

    private static List<Character> merge(List<Character> left, List<Character> right) {
        List<Character> merged = new ArrayList<Character>(left.size() + right.size());
        int i = 0;
        int j = 0;
        while (i < left.size() && j < right.size()) {
            char leftLetter = left.get(i);
            char rightLetter = right.get(j);
            int comparison = Character.compare(leftLetter, rightLetter);
            if (comparison < 0) {
                merged.add(leftLetter);
                ++i;
            } else if (comparison > 0) {
                merged.add(rightLetter);
                ++j;
            } else {
                // letters are same, add both
                merged.add(leftLetter);
                ++i;
                merged.add(rightLetter);
                ++j;
            }
        }
        while(i < left.size()) {
            merged.add(left.get(i));
            ++i;
        }
        while(j < right.size()) {
            merged.add(right.get(j));
            ++j;
        }
        return merged;
    }
} 

Answer 4

Managed to get it working. As its a very specific thing I'm not sure how much use it will be but the general case is as explained above.

for(int k = 0, i = 0; k < decryptedtextColumns.get(i%5).size(); k++)
    {           
        i = 0;
        while(i < decryptedtextColumns.size() && k < decryptedtextColumns.get(i).size())
        {
            if(k <= decryptedtextColumns.get(i).size())
            {
                Character letter = decryptedtextColumns.get(i).get(k);
                decryptedtext.add(letter);
            }
            i++;
        }
    }