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:

...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?

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