【leetcode】419.Battleships in a Board

原題

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

求戰艦數
一個矩陣，.表示空，X表示戰艦
戰艦隻能橫着或豎着放，且不會挨着，至少相隔一個位置
求矩陣中戰艦的數量

思路

從左到右，從上到下遍歷判斷，找到一個左上沒有相連的X戰艦數+1，其餘的X均是戰艦的一部分

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;
    }