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:

Example:

X..X...X...XIn the above board there are 2 battleships.
Invalid Example:
...XXXXX...X
This is an invalid board that you will not receive - as battleships will always have a cell separating between them.
思路：
簡單的思路是做搜索。但是考慮到每個船和每個船之間至少橫向縱向有一個空格，那么如果只考慮每艘船的右下角：
XX.X...X
這樣的話，右下角的X右邊和下面都是空格，每次碰到符合這個條件的X就認為碰到一個船，否則不管。
題解：
int countBattleships(const std::vector<std::vector<char>>& board) {    const int M = board.size();    const int N = board[0].size();    int numShips(0);    for(int i = 0; i < M; ++i) {        for(int j = 0; j < N; ++j) {            if (board[i][j] == 'X') {                numShips += ((i < M - 1 && board[i + 1][j] == '.') || (i == M - 1)) &&                            ((j < N - 1 && board[i][j + 1] == '.') || (j == N - 1));            }        }    }    return numShips;}