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?

class Solution:
    def countBattleships(self, board):
        """
        :type board: List[List[str]]
        :rtype: int
        """
        res = 0
        m,n = len(board),len(board[0])
        for i in range(m):
            for j in range(n):
                if board[i][j]=='X':
                    if not ((i-1>=0 and board[i-1][j]=='X') or (j-1>=0 and board[i][j-1]=='X')):
                        res += 1
        return res