419. Battleships in a Boardv-优快云博客

本文链接：https://blog.youkuaiyun.com/wilzxu/article/details/90253431

本文介绍了一种在二维棋盘上计数战舰数量的算法。通过检查每个战舰的起点，即最左点和最上点，实现O(mn)时间复杂度和O(1)空间复杂度的解决方案。

419. Battleships in a Board

方法1:
- Complexity

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.

Follow up:

Could you do it in one-pass, using only O(1) extra memory and without modifying the value of the board?

方法1:

思路：

由于题目中限定了不会出现连接的行和列，只要计数每一个battleship的“起点”即可。所谓起点，就是最左点和最上点。也就是说，仅当x出现在左边界或者上边界或者左点和上点均不为“.“的时候才参与计数。

注意这道题的叙述比较坑，并没有让你判断board是否合法，只是计数，题目本身保证了不会出现不合法题目。所以用最正常的dfs/bfs就能算出来，只不过只有加上这个限制才能使O(1)空间可能。

Complexity

Time complexity: O(mn)
Space complexity: O(1)

class Solution {
public:
    int countBattleships(vector<vector<char>>& board) {
        int m = board.size(), n = board[0].size(), result = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (board[i][j] == '.') continue;
                if (i > 0 && board[i - 1][j] == 'X') continue;
                if (j > 0 && board[i][j - 1] == 'X') continue;
                result++;
            }
        }
        return result;
    }
};