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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0401_0500.s0419_battleships_in_a_board;

// #Medium #Array #Depth_First_Search #Matrix #2022_07_16_Time_0_ms_(100.00%)_Space_41.9_MB_(88.66%)

/**
 * 419 - Battleships in a Board\.
 *
 * Medium
 *
 * Given an `m x n` matrix `board` where each cell is a battleship `'X'` or empty `'.'`, return _the number of the **battleships** on_ `board`.
 *
 * **Battleships** can only be placed horizontally or vertically on `board`. In other words, they can only be made of the shape `1 x k` (`1` row, `k` columns) or `k x 1` (`k` rows, `1` column), where `k` can be of any size. At least one horizontal or vertical cell separates between two battleships (i.e., there are no adjacent battleships).
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/04/10/battelship-grid.jpg)
 *
 * **Input:** board = \[\["X",".",".","X"],[".",".",".","X"],[".",".",".","X"]]
 *
 * **Output:** 2 
 *
 * **Example 2:**
 *
 * **Input:** board = \[\["."]]
 *
 * **Output:** 0 
 *
 * **Constraints:**
 *
 * *   `m == board.length`
 * *   `n == board[i].length`
 * *   `1 <= m, n <= 200`
 * *   `board[i][j]` is either `'.'` or `'X'`.
 *
 * **Follow up:** Could you do it in one-pass, using only `O(1)` extra memory and without modifying the values `board`?
**/
public class Solution {
    public int countBattleships(char[][] board) {
        if (board == null || board.length == 0) {
            return 0;
        }
        int count = 0;
        int m = board.length;
        int n = board[0].length;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (board[i][j] != '.'
                        && (j <= 0 || board[i][j - 1] != 'X')
                        && (i <= 0 || board[i - 1][j] != 'X')) {
                    count++;
                }
            }
        }
        return count;
    }
}