LeetCode #419: Battleships in a Board

Problem Statement

(Source) 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.

Solution

从左向右，从上向下遍历board，只需要判断当前所在位置是否为

X

以及是否能使得battership的数量增加，判断逻辑见如下代码。算法的时间复杂度为board的cell数量。

class Solution(object):
def countBattleships(self, board):
"""
:type board: List[List[str]]
:rtype: int
"""
res = 0
n, m = len(board), len(board[0])
for i in xrange(n):
for j in xrange(m):
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