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.

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

按道理一般情况下follow up的要求是满足不了的，但是由于battleship的形状是固定的，只能是直线，所以可以满足要求

class Solution {
    public int countBattleships(char[][] board) {
        if(board == null || board.length == 0 || board[0] == null || board[0].length == 0)
            return 0;

        int res = 0;
        int[] last = null;
        int m = board.length;
        int n = board[0].length;

        for(int i = 0; i < m; i++) {
            for(int j = 0; j < n; j++) {
                if(board[i][j] == 'X') {
                    if(i - 1 >= 0 && board[i - 1][j] == 'X')
                        continue;
                    if(j - 1 >= 0 && board[i][j - 1] == 'X')
                        continue;
                    res++;
                }
            }
        }

        return res;
    }
}