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?

Solution:

class Solution(object):
    def countBattleships(self, board):
        """
        :type board: List[List[str]]
        :rtype: int
        """
        ships = 0
        for i, row in enumerate(board):
            for j, val in enumerate(row):
                if val != 'X':
                    continue
                if i > 0 and board[i - 1][j] == 'X':
                    continue
                if j > 0 and board[i][j - 1] == 'X':
                    continue
                ships += 1
        return ships