题目说明：

Given an 2D board, count how many different 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.

解法1思路：

这里需要计算战舰队的数量count，可转换为计算每个战舰队位于最左上角的船只（我们称为战舰队头）。 用两个循环，对网格内的每一个格子进行遍历，若该格子为X（即有战舰），则分以下几种情况： 
若在网格的最左上角（即i=0，j=0），则存在一个舰队，count自加1； 
若在网格的最上角（即i=0），但不在最左边（即j!=0），则需要判断该位置的左边（即board[i][j-1]）是否有战舰存在，若其左边不存在战舰，则该位置为战舰队头，count自加1； 
若在网格的最左边（即j=0），但不在最上边（即i!=0），则需要判断该位置的上面（即board[i-1][j]）是否有战舰存在，若其上面不存在战舰，则该位置为战舰队头，count自加1； 
若战舰不在最左边也不在最上边（即i!=0且j!=0），则需要判断该战舰的左边和上面是否还有战舰存在，若其左边和上面都不存在战舰，则该位置为战舰队头，count自加1； 
其余情况，战舰都不属于战舰队头。

解法1代码：

int countBattleships(vector<vector<char>>& board)
{
    int counter = 0;
    for (int i = 0; i < board.size(); i ++) {
        for (int j = 0; j < board[i].size(); j ++) {
            if (board[i][j] == ‘X‘) {
                if (i == 0 && j == 0)
                    counter ++;
                else if (i == 0 && j != 0 && board[i][j - 1] == ‘.‘)
                    counter ++;
                else if (j == 0 && i != 0 && board[i - 1][j] == ‘.‘)
                    counter ++;
                else if (i != 0 && j != 0 && board[i - 1][j] == ‘.‘ && board[i][j - 1] == ‘.‘)
                    counter ++;
            }
            
        }
    }

    return counter;
}

解法2思路：

和解法1相似，只是这种解法是从反面分析，逐项排除非battleships的项，最后留下符合条件的项，这种剪枝策励在leetcode的算法题中非常常见。

解法2代码：

int countBattleships(vector<vector<char>>& board)
{
    int counter = 0;
    for (int i = 0; i < board.size(); i ++) {
        for (int j = 0; j < board[i].size(); j ++) {
            if (board[i][j] == ‘.‘)
                continue;
            if (j > 0 && board[i][j - 1] == ‘X‘)
                continue;
            if (i > 0 && board[i - 1][j] == ‘X‘)
                continue;
            
            ++ counter;
        }
    }
    
    return counter;
}

部分引用自：

http://blog.5ibc.net/p/94878.html

和

http://blog.csdn.net/mebiuw/article/details/52876700

leetcode 419