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