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?

class Solution {
    public int countBattleships(char[][] board) {
        int m = board.length;
        if(m == 0) return 0;
        int n = board[0].length;
        int res = 0;
        
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(board[i][j] == '.') continue;
                if(i > 0 && board[i-1][j] == 'X') continue;
                if(j > 0 && board[i][j-1] == 'X') continue;
                res++;
            }
        }
        return res;
    }
}

牛逼啊，只count左上角是x的（潜在的ship头）如果当前x的左边或上面还是x就要continue，否则就是一个新的ship