python--leetcode 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:

...XXXXX...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?

计算战舰数量，战舰有可能横着也有可能竖着。这一题的关键就是战舰之间一定有‘.’来隔开。

因此我们可以得到一个结论：每一艘战舰，有且只有一个‘X’的前一列以及前一行都为‘.’

代码如下：

class Solution(object):    def countBattleships(self, board):        if len(board) == 0: return 0        m, n = len(board), len(board[0])        count = 0        for i in range(m):            for j in range(n):                if board[i][j] == 'X' and (i == 0 or board[i-1][j] == '.') and (j == 0 or board[i][j-1] == '.'):                    count += 1        return count