419. Battleships in a Board

题目描述：

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.

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.

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

解题思路：

善于利用题目信息”This is an invalid board that you will not receive - as battleships will always have a cell separating between them.“，battleship不相邻，一个点如果是battleship中一点，当battleship的长度大于一时其横向或者纵向临近的点肯定是'X'，当battleship的长度为一时其临近的点都是‘.’。因此，对于每个battleship，我们只统计其纵向或者横向的第一个点。

代码：

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