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:

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X..X
...X
...X

Invalid Example:

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...X
XXXX
...X

Follow up:
Could you do it in one-pass, using only O(1) extra memory and without modifying the value of the board?

思路

Didn’t work it out at first.

The initial consideration is to construct a visited (array[]). If visited and search downward and rightward, shall be remembered in this array. But the memory is not O(1), but O(n)

So this method comes quite handy and simple. Remember that the Online Judge system doesn’t require the complexity of algoritm to be strictly less than O(n^2), so feel free to try the brute force method at first.

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class Solution(object):
def countBattleships(self, board):
"""
:type board: List[List[str]]
:rtype: int
"""
lenlines = len(board)
lencols = len(board[0])
count = 0
for i in range(lenlines):
for j in range(lencols):
if board[i][j] == "X":
preleft = False
preup = False
if i - 1 >= 0:
if board[i-1][j] == "X":
preup = True
if j - 1 >= 0:
if board[i][j-1] == "X":
preleft = True
if not (preup or preleft):
count += 1
return count