LeetCode 419. Battleships in a Board （算法）

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?

输入一组由‘X’和‘.’组成的矩阵，X代表船，1xN或Nx1艘船能组成一个舰队，舰队互不相邻（由点隔开），求舰队数量。

思路：对于一个X点，如果其上方和左侧都不是X，则表明该点是舰队的边缘，计数加1。注意第一排和第一列的处理（如果是第一排，则只检查左侧；如果是第一列，则只检查上方）。

    int countBattleships(vector<vector<char>>& board) {        int i,j,count=0;        for(i=0;i<board.size();i++)        {            for(j=0;j<board[0].size();j++)            {                if(board[i][j]=='X'&&(i==0||board[i-1][j]!='X')&&(j==0||board[i][j-1]!='X'))                    count++;            }        }        return count;    }