【leetcode】419. Battleships in a Board【E】

Given an 2D board, count how many different 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 not a valid board - as battleships will always have a cell separating between them.

Your algorithm should not modify the value of the board.

Subscribe to see which companies asked this question

程序写的好长

不知道为啥这算是个E还过了这么多，估计没准儿自己程序写的不咋好。。。

分别横着竖着，如果一个船旁边儿有别的船，那就不是

class Solution(object):
def countBattleships(self, board):

bb = board
m = len(board)
n = len(board[0])

b = [[u'.']*(n+2)]
for i in bb:
tmp = ([u'.'] + i + [u'.'])
b += tmp[:],
b += [u'.']*(n+2),

i,j = 1,1
res = 0
while i < (m+1):
j = 0
while j < (n+1):
flag = False
while j < n+1 and b[i][j] == 'X':
if b[i-1][j] == 'X' or b[i+1][j] == 'X':
flag = False
break
b[i][j] = '.'
flag = True
j += 1
if flag:
res += 1
j += 1
i += 1

i = 1
j = 1

while j < n+1:
i = 1
while i < m+1:
flag = False
while i < m+1 and b[i][j] == 'X':
if b[i][j+1] == 'X' or b[i][j-1] == 'X':
flag = False
break
b[i][j] = '.'
flag = True
i += 1
if flag:
res += 1

i += 1
j += 1

return res