leetcode 419

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 = ;
    for (int i = ; i < board.size(); i ++) {
        for (int j = ; j < board[i].size(); j ++) {
            if (board[i][j] == 'X') {
                if (i ==  && j == )
                    counter ++;
                else if (i ==  && j !=  && board[i][j - ] == '.')
                    counter ++;
                else if (j ==  && i !=  && board[i - ][j] == '.')
                    counter ++;
                else if (i !=  && j !=  && board[i - ][j] == '.' && board[i][j - ] == '.')
                    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 = ;
    for (int i = ; i < board.size(); i ++) {
        for (int j = ; j < board[i].size(); j ++) {
            if (board[i][j] == '.')
                continue;
            if (j >  && board[i][j - ] == 'X')
                continue;
            if (i >  && board[i - ][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