【leetcode】419.Battleships in a Board

原题

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?

解析

求战舰数
一个矩阵，.表示空，X表示战舰
战舰只能横着或竖着放，且不会挨着，至少相隔一个位置
求矩阵中战舰的数量

思路

从左到右，从上到下遍历判断，找到一个左上没有相连的X战舰数+1，其他的X均是战舰的一部分

解法

public int countBattleships(char[][] board) {
        int count = 0;
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                if (board[i][j] == 'X') {
                    if ((i > 0 && board[i - 1][j] == 'X') || j > 0 && board[i][j - 1] == 'X') {
                        continue;
                    }
                    count++;
                }
            }
        }
        return count;
    }