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?

Solution: 找到battleship的起点和终点。由于battleship只能横放或竖放，所以battleship的上边或左边必须为'.'。

public class Solution {
    public int countBattleships(char[][] board) {
        if (board == null || board[0] == null) return 0;
        
        int m = board.length, n = board[0].length;
        int result = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (board[i][j] == '.' || (i > 0 && board[i - 1][j] == 'X') || (j > 0 && board[i][j - 1] == 'X'))
                    continue;
                else result++;
            }
        }
        return result;
    }
}