LeetCode题解：Battleships in a Board

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; }