[LeetCode]419. Battleships in a

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) 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?

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