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.

找到有多少条船，横向或纵向不相隔的X组成一条船，船之间至少有一个空点

解题思路：

遍历二维数组，每遍历到一个位置的时候，只需要考虑其左边和上面的位置是否为X即可

 int countBattleships(char** board, int boardRowSize, int boardColSize) {

     int i,j;

     int count=;

     for(i=;i<boardRowSize;i++)

         for(j=;j<boardColSize;j++)

             if(board[i][j]=='X')

             {

                 if(i==)

                 {

                     if(j==||board[i][j-]=='.')

                         count++;

                 }

                 else

                 {

                     if((j==&&board[i-][j]=='.')||(board[i-][j]=='.'&&board[i][j-]=='.'))

                     count++;

                 }

             }

             return count;

 }

 //遍历到每个元素时候，只需要考虑其前面和上面是否为'.'即可