LeetCode题解：Battleships in a Board_given an 2d board, count how many battleships are -优快云博客

本文介绍了一种简单有效的算法来计数二维棋盘上的战舰数量。通过搜索每艘战舰的右下角，判断其是否为独立战舰，以此实现高效计数。文章提供了完整的代码实现。

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.

思路：

简单的思路是做搜索。但是考虑到每个船和每个船之间至少横向纵向有一个空格，那么如果只考虑每艘船的右下角：

XX.X
...X

这样的话，右下角的X右边和下面都是空格，每次碰到符合这个条件的X就认为碰到一个船，否则不管。

题解：

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;
}