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.

``` 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```