创建一个N x N的字母网格，以便没有两个相邻的字母相同

Question

I need to print out a N by N grid containing the letters A to F, so that no two adjacent letters are the same. The code below prints out an N by N grid, however I can only get the letters on the left and right to be different. I can't find a way to get the letters above and below to be different as well. I need to solve this problem without the use of arrays. The letters have to be randomized.

public static void main(String[] args) {
    int N = StdIn.readInt();
    for (int column = 0; column < N; column++) {
        int x = 0;

        for (int row = 0; row < N; row++) {

            int c = (int) (Math.random() * 6 + 1);

            while (x == c) {
                c = (int) (Math.random() * 6 + 1);
            }
            if (c == 1) {
                System.out.print("A ");
            }
            if (c == 2) {
                System.out.print("B ");
            }
            if (c == 3) {
                System.out.print("C ");
            }
            if (c == 4) {
                System.out.print("D ");
            }
            if (c == 5) {
                System.out.print("E ");
            }
            if (c == 6) {
                System.out.print("F ");
            }

            x = c;

        }

        System.out.println();

    }

Answer 1

I did you the favor of simplifying your for loop a bit

for (int row = 0; row < N; row++) {
  char c = (char) (Math.random() * 6 + 'A');
  while (x == c) {
    c = (char) (Math.random() * 6 + 'A');
  }
  System.out.print(c + " ");
  x = c;
}

That uses the ASCII values of the letters so you don't need a big if statement.

Your question is unclear about what storage formats are allowed, but consider this: If you could store each row as a string (one temporary string that is deleted after moving on a row), how would you check if a letter in the row you're building below it matches the letter above? Think of a string as an array (which is its true form).

Answer 2

Easy to do with 2 queues.

If you analyse the below, you should realise that you really only need one size N linked-list.

current is the current row, each new element is simply enqueued.

last is the previous row. It takes the value of current once you're done with a row, then the first element we want to check is at the front.

I separated the first row from the others to make things simpler and more understandable.

  int N = 10;
  Queue<Character> last,
                   current = new Queue<Character>();
  char prev = '0';
  for (int row = 0; row < N; row++)
  {
     char c;
     do { c = (char)(Math.random() * 6 + 'A'); }
     while (prev == c);
     current.enqueue(c);
     prev = c;
     System.out.print(c + " ");
  }
  System.out.println();
  for (int col = 1; col < N; col++)
  {
     last = current;
     current = new Queue<Character>();
     prev = '0';
     for (int row = 0; row < N; row++)
     {
        char c;
        do { c = (char)(Math.random() * 6 + 'A'); }
        while (last.peek() == c || prev == c);
        current.enqueue(c);
        last.dequeue();
        prev = c;
        System.out.print(c + " ");
     }
     System.out.println();
  }

Answer 3

You can do this with a pair of stacks and a candidate character. The NxN grid is represented by a stack containing up to N times N characters. Try this with a deck of cards, with the stacks represented by two spots for stacks of cards, to get a feel for how it works.

1. Create two stacks, A and B. Keep a running index for the number of items in A.
2. Generate a random number. This is your Candidate.
3. If there are a multiple of N items in stack A (you just finished a row) skip to step 6.
4. Pop the top of stack A and store it as Check.
5. Is Check adjacent to Candidate? If yes, push Check back onto stack A, and go back to step 2. (null is non-adjacent)
6. Push Check onto stack B. Pop N-2 more items, pushing them onto stack B as you go.
7. Pop the top of stack A and store it as Check. (There should be N-1 items in stack B, and one in Check: a whole row)
8. Is Check adjacent to Candidate? (null is non-adjacent) If yes, push Check back onto stack A, pop each of the items from stack B, pushing onto stack A as you go, and go back to step 2.
9. push check back onto stack A. Pop each item from stack B, pushing onto stack A as you go.
10. Push Candidate onto stack A.
11. Increment the index. If we now have NxN items in stack A, we're done: pop everything off stack A, writing them down in an NxN grid as you go. Otherwise return to step 2.

Basically, you're drawing a card, and checking up to two items. Because stack A is guaranteed never to have adjacencies in it, once an item is placed, it never has to be removed. So once you get to the end, you're done.