Battleships in a Board问题及解法

问题描述：

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.

示例:

X..X...X...X

In the above board there are 2 battleships.

无效示例:

...XXXXX...X

This is an invalid board that you will not receive - as battleships will always have a cell separating between them.

问题分析：

每次遍历一行，遍历到的值为‘X’，并且该值若为某个battleship的第一个，count++。

过程详见代码：

class Solution {public:    int countBattleships(vector<vector<char>>& board) {        int m = board.size();if (!m) return 0;int n = board[0].size();int count = 0;for (int i = 0; i < m; i++){for (int j = 0; j < n; j++){if (board[i][j] == '.') continue;if (i > 0 && board[i - 1][j] == 'X') continue;if (j > 0 && board[i][j - 1] == 'X') continue;count++;}}return count;    }};