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

 1 int countBattleships(char** board, int boardRowSize, int boardColSize) {
 2     int i,j;
 3     int count=0;
 4     for(i=0;i<boardRowSize;i++)
 5         for(j=0;j<boardColSize;j++)
 6             if(board[i][j]==‘X‘)
 7             {
 8                 if(i==0)
 9                 {
10                     if(j==0||board[i][j-1]==‘.‘)
11                         count++;
12                 }
13                 else
14                 {
15                     if((j==0&&board[i-1][j]==‘.‘)||(board[i-1][j]==‘.‘&&board[i][j-1]==‘.‘))
16                     count++;
17                 }
18             }
19             return count;
20 }
21 //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

419. Battleships in a Board