Given an m x n matrix board where each cell is a battleship 'X' or empty '.', return the number of the battleships on board.
Battleships can only be placed horizontally or vertically on board. In other words, they can only be made of the shape 1 x k (1 row, k columns) or k x 1 (k rows, 1 column), where k can be of any size. At least one horizontal or vertical cell separates between two battleships (i.e., there are no adjacent battleships).
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Example 1:
\n![\"\"](\"https://assets.leetcode.com/uploads/2021/04/10/battelship-grid.jpg\")
\n\nInput: board = [["X",".",".","X"],[".",".",".","X"],[".",".",".","X"]]\nOutput: 2\n\n\n
Example 2:
\n\n\nInput: board = [["."]]\nOutput: 0\n\n\n
\n
Constraints:
\n\nm == board.lengthn == board[i].length1 <= m, n <= 200board[i][j] is either '.' or 'X'.\n
Follow up: Could you do it in one-pass, using only O(1) extra memory and without modifying the values board?
给你一个大小为 m x n 的矩阵 board 表示甲板，其中，每个单元格可以是一艘战舰 'X' 或者是一个空位 '.' ，返回在甲板 board 上放置的 战舰 的数量。
战舰 只能水平或者垂直放置在 board 上。换句话说，战舰只能按 1 x k（1 行，k 列）或 k x 1（k 行，1 列）的形状建造，其中 k 可以是任意大小。两艘战舰之间至少有一个水平或垂直的空位分隔 （即没有相邻的战舰）。
\n\n
示例 1：
\n\n\n输入：board = [[\"X\",\".\",\".\",\"X\"],[\".\",\".\",\".\",\"X\"],[\".\",\".\",\".\",\"X\"]]\n输出：2\n\n\n
示例 2：
\n\n\n输入：board = [[\".\"]]\n输出：0\n\n\n
\n\n
提示：
\n\nm == board.lengthn == board[i].length1 <= m, n <= 200board[i][j] 是 '.' 或 'X'\n\n
进阶：你可以实现一次扫描算法，并只使用 O(1) 额外空间，并且不修改 board 的值来解决这个问题吗？
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