LeetCode: Battleships in a Board

// 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:
// X..X
// ...X
// ...X
// In the above board there are 2 battleships.

// Invalid Example:
// ...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?

public class BattleshipsInABoard {
    public int countBattleships(char[][] board) {
        int ships = 0;
     
        for(int i = 0; i < board.length; i++) {
            for(int j = 0; j < board[0].length; j++) {
                if(board[i][j] == 'X') {
                    ships++;
                    sink(board, i, j, 1);
                }
            }
        }
     
        return ships;
    }
 
    public void sink(char[][] board, int i, int j, int numberOfShips) {
        if(i < 0 || i >= board.length || j < 0 || j >= board[0].length || board[i][j] == '.') {
            return;
        }
     
        board[i][j] = '.';
        sink(board, i + 1, j, numberOfShips + 1);
        sink(board, i, j + 1, numberOfShips + 1);
    }
}