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?

写的和答案基本一致

从左上到右下 如果’X’的左边或上方有’X’ 那么忽略 不做重复计算 

    public int countBattleships(char[][] board) {        int m = board.length;        if (m==0) return 0;        int n = board[0].length;                int count=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;                count++;            }        }                return count;    }