题目
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) orNx1(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 inone-pass, using onlyO(1) extra memoryandwithout modifyingthe value of the board?
难度
Medium
方法
遍历board，对于board[i][j]=='X'，如果board[i-1][j]或者board[i][j-1]不为'X'(i-1和j-1均得>=0)，则board[i][j]=='X'为一个battleship
python代码
class Solution(object):
def countBattleships(self, board):
"""
:type board: List[List[str]]
:rtype: int
"""
count = 0
for i in range(len(board)):
for j in range(len(board[0])):
if board[i][j]=='X' and ((i > 0 and board[i-1][j]!='X') or i==0) and ((j > 0 and board[i][j-1]!='X') or j==0):
count += 1
return count
assert Solution().countBattleships(["X..X","...X","...X"]) == 2
assert Solution().countBattleships(["...X","XXXX","...X"]) == 2
