[LeetCode] Problem 419 - Battleships in a Board

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Given an 2D board, count how many battleships are in it. The battleships are represented with ‘X’s, empty slots are represented with ‘.’s. You may assume the following rules:

  • You receive a valid board, made of only battleships or empty slots.
  • Battleships can only be placed horizontally or vertically. In other words, they can only be made of the shape 1xN (1 row, N columns) or Nx1 (N rows, 1 column), where N can be of any size.
  • At least one horizontal or vertical cell separates between two battleships - there are no adjacent battleships.

Example:

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X..X
...X
...X

In the above board there are 2 battleships.

Invalid Example:

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...X
XXXX
...X

This is an invalid board that you will not receive - as battleships will always have a cell separating between them.

Follow up

Could you do it in one-pass, using only O(1) extra memory and without modifying the value of the board?

Code

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public int countBattleships(char[][] board) {
    int count = 0;

    for (int i = 0; i < board.length; i++) {
        for (int j = 0; j < board[0].length; j++) {
            if (board[i][j] == 'X' && (i == 0 || board[i - 1][j] == '.') && 
                                        (j == 0 || board[i][j - 1] == '.'))
                count++;
        }
    }

    return count;
}