算法设计作业13

第十三周作业：

419. Battleships in a Board

解题思路：

419. Battleships in a Board

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.

思路：给一个甲板，判断有几艘战舰。战舰的特点是，横着的或者竖着的一个或连续多个的'X'，且两艘战舰不能相邻。

由于战舰只能横着或者竖着，所以如果从第一行开始从左向右遍历的话，只需判断一个新的战舰要么是在最左或最上，要么就是它的左边和上边的元素不是'X'。所以只用遍历一次就可以完成整个计算。

代码如下：

class Solution {
public:
    int countBattleships(vector>& board) {
        if (board.empty() || board[0].empty()) {
            return 0;
        }

        int cnt = 0;
        for (int i = 0; i < board.size(); ++i) {
            for (int j = 0; j < board[0].size(); ++j) {
                cnt += board[i][j] == 'X' &&
                       (i == 0 || board[i - 1][j] != 'X') &&
                       (j == 0 || board[i][j - 1] != 'X');
            }
        }
        return cnt;
    }
};