419. Battleships in a Board

DFS

T: O(mn)

class Solution {
    private static int[][] DIRS = {{1,0}, {0, 1}, {-1,0}, {0,-1}};
    public int countBattleships(char[][] board) {
        int m = board.length;
        int n = board[0].length;
        
        int count = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (board[i][j] == 'X') {
                    dfs(board, i, j);
                    count++;
                }
            }
        }
        return count;
    }
    private void dfs(char[][] board, int i, int j) {
        int m = board.length;
        int n = board[0].length;
        if (!(i >= 0 && i < m && j >= 0 && j < n) || board[i][j] != 'X') {
            return;
        }
        board[i][j] = '.';
        for (int[] dir : DIRS) {
            dfs(board, i + dir[0], j + dir[1]);
        }
        
    }
}

Follow up: Could you do it in one-pass, using only O(1) extra memory and without modifying the values board?

class Solution {
    public int countBattleships(char[][] board) {
        int m = board.length;
        int n = board[0].length;
        
        int count = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                // meets this condition, it must a head of a ship 
                if (board[i][j] == 'X' && (i == 0 || board[i-1][j] == '.') && (j == 0 || board[i][j-1] == '.')) {
                    count++;
                }
            }
        }
        return count;
    }
}