419. Battleships in a Board 难度：medium

最新推荐文章于 2021-12-18 20:46:20 发布

luoyingmin

最新推荐文章于 2021-12-18 20:46:20 发布

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本文介绍了一种用于计数二维棋盘中战舰数量的算法。战舰由字符 'X' 表示，空位由 '.' 表示。通过遍历棋盘并检查每个 'X' 是否能增加战舰计数来实现。确保了战舰之间至少有一个空位相隔。

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题目：

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.

思路：

从左向右，从上向下遍历board，只需要判断当前所在位置是否为X 以及是否能使得battership的数量增加

程序：

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