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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本文介绍了一个算法问题，即如何在一给定的二维棋盘中计算战舰的数量。战舰由字符'X'表示，空位则用'.'表示。文章提供了一个解决方案，通过遍历棋盘来寻找每个战舰的起始位置，确保不重复计数。

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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?

因为X只会成一行或者一列粗线，并且两两之间一定会分隔开

那么只要计算一行的最左边一个和一列的最上面一个就行

public class Solution {
    public int countBattleships(char[][] board) {
          if(board == null || board.length == 0){
              return 0;
          }
          
          int count = 0;
          for(int i = 0; i < board.length; i++){
              for(int j = 0; j < board[0].length; j++){
                  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;
    }
}