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:

...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?

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：

1 class Solution { 2 public: 3     int countBattleships(vector
     
      
       >& board) { 4         int num = 0; 5         int col = board[0].size(); 6         int row = board.size(); 7         for (int i = 0; i < row; ++i) { 8             for (int j = 0; j < col; ++j) { 9                 if (board[i][j] == 'X') {10                     if (i > 0 && board[i-1][j] == 'X')11                         continue;12                     if (j > 0 && board[i][j-1] == 'X')13                         continue;14                     num++;15                 }16             }17         }18         return num;19     }20 };