419. Battleships in a Board

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?

题目要求统计矩阵中表示的有几艘战舰。这里战舰是不能相连的，所以可以简单根据有多少个头部来计算，一个‘X‘点的上方和左方不是‘X’说明这是一艘船的头。

代码：

class Solution
{
public:
	int countBattleships(vector<vector<char> >& board)
	{
		int m = board.size();
		if(m == 0) return 0;
		int n = board[0].size();
		if(n == 0) return 0;

		int res = 0;

		for(int i = 0; i < m; ++i)
		{
			for(int j = 0; j < n; ++j)
			{
				if(board[i][j] == 'X' && (i == 0 || board[i - 1][j] != 'X') && (j == 0 || board[i][j - 1] != 'X'))
				{
					++res;
				}
			}
		} 
		return res;
	}
};