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 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?

 public class Solution {
public int countBattleships(char[][] board) {
int m = board.length;
if (m == ) return ;
int n = board[].length, count = ; for (int i = ; i < m; i++) {
for (int j = ; j < n; j++) {
if (board[i][j] == '.') continue;
if (i > && board[i - ][j] == 'X') continue;
if (j > && board[i][j - ] == 'X') continue;
count++;
}
}
return count;
}
}

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