題目：

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?

解釋：
只找表示船頭的元素個數就好，這種思路利用了船隻能在水平方向或是垂直方向上延伸，因此船的最左邊（水平方向）或是最上邊（垂直方向）的元素就很特殊，因為它的左邊（水平方向）或是上邊（垂直方向）的元素會是’.’，而對於船中部的元素，它們的左邊或是上邊的元素肯定有一個會是’X’。
利用這點，我們可以遍歷矩陣每一個元素，當符合是船頭元素時，就增加船的個數，否則繼續遍歷下一個元素，最後返回找到的船的個數即可。
python程式碼：

class Solution(object):
    def countBattleships(self, board):
        """
        :type board: List[List[str]]
        :rtype: int
        """
 

        m=len(board)
        n=len(board[0])
        count=0
        for i in xrange(m):
            for j in xrange(n):
                if board[i][j]=='X' and (i==0 or board[i-1][j]=='.') and (j==0 or board[i][j-1]=='.'):
                    count+=1
        return count       

c++程式碼：

class Solution {
public:
    int countBattleships(vector<vector<char>>& board) {
        int cnt=0;
        for (int i=0;i<board.size();i++)
        {
            for(int j=0;j<board[0].size();j++)
            {
                if(board[i][j]=='X' &&(i==0 || board[i-1][j]=='.') &&(j==0 ||board[i][j-1]=='.'))
                    cnt++;
            }
        }
        return cnt;
    }
};

總結：
要學會合理地轉化問題。