419. 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?

思路：
题目进阶要求一次遍历完，并且使用O(1)的空间复杂度，那么就不能像解法一那样再声明一个记录访问标志的二维数组。我们可以通过军舰的起点计算军舰数，所谓起点，就是指一条军舰上最左边的那个‘X’或者最上面的那个‘X’。

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