[leetcode]419. Battleships in a Board

最新推荐文章于 2024-07-24 17:53:27 发布

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最新推荐文章于 2024-07-24 17:53:27 发布

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本文介绍了一个算法问题：如何在一给定的有效二维棋盘中计算战舰的数量。战舰由字符 'X' 表示，空位则为 '.'。战舰只能水平或垂直放置，并且彼此间至少有一个空位隔开。文章提供了详细的解决方案，通过遍历矩阵并检查每个可能的战舰起点来计算战舰总数。

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Question: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?

Analysis:这个题，就是用矩阵一个放战舰的地方，X是放的地方，.是空槽

他肯定给成横着的一条或者竖着的一条，而且肯定有隔离，不会交叉。。现在求出有多少个战舰

因为给定的数据肯定是有效的，所以我们直接统计最左上的节点的数量就好，就是每一个战舰的最最上角，统计一下数额就是了

Answer:

public class Solution {
    public int countBattleships(char[][] board) {
         int m = board.length;
        if (m==0) return 0;
        int n = board[0].length;
        int count=0;

        for (int i=0; i<m; i++) {
            for (int j=0; j<n; j++) {
                //找到最top-left的那一个，并且判断其是否和合法，也就是没有交叉
                if (board[i][j] == '.') continue;
                if (i > 0 && board[i-1][j] == 'X') continue;
                if (j > 0 && board[i][j-1] == 'X') continue;
                count++;
            }
        }

        return count;
    }
}