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?

 1 public class Solution {
 2     public int countBattleships(char[][] board) {
 3         int m = board.length;
 4         if (m == 0) return 0;
 5         int n = board[0].length, count = 0;
 6 
 7         for (int i = 0; i < m; i++) {
 8             for (int j = 0; j < n; j++) {
 9                 if (board[i][j] == '.') continue;
10                 if (i > 0 && board[i - 1][j] == 'X') continue;
11                 if (j > 0 && board[i][j - 1] == 'X') continue;
12                 count++;
13             }
14         }
15         return count;
16     }
17 }