LeetCode-Battleships in a Board

1. Battleships in a Board（Medium）

Description
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:

...XXXXX...X

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

Follow up:
Could you do it in one-pass, using only O(1) extra memory and without modifying the value of the board?

Analysis
分析题意，题目要求我们找出船的数量，船只有水平和垂直两种形式，并且题目规定两条船不能相互接触，所以我们找到每条船的船头就可以知道所有船的数量。即找到所有的上方和左方没有'X'的'X'。

代码：

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