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?

本题题意很简单，最简单的方法就是直接遍历求解，修改棋盘即可，但是我在网上看到了一个更加简单的方法，思路也很新奇，所以就这么做了，思想的本质就是按照行统计数量

代码如下：

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <cmath>

using namespace std;

class Solution 
{
public:
    int countBattleships(vector<vector<char>>& board)
    {
        if (board.size() <= 0)
            return 0;
        int row = board.size(), col = board[0].size();
        int count = 0;
        for (int i = 0; i < row; i++)
        {
            for (int j = 0; j < col; 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;
    }
};