419. Battleships in a Board

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?

int countBattleships(vector<vector<char>>& board) { // time: O(m * n); space: O(1)
    int m = board.size(), n = board[0].size(), res = 0;
    for (int i = 0; i < m; ++i) {
        for (int j = 0; j < n; ++j) {
            if (board[i][j] == '.' || 
                (i > 0 && board[i - 1][j] == 'X') || 
                (j > 0 && board[i][j - 1] == 'X')) continue; 
            ++res;
        }
    }
    return res;
}