_419.java

package com.fishercoder.solutions;

/**
 * 419. Battleships in a Board
 *
 * 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?

 */
public class _419 {

    public static class Solution1 {
        /**
         * credit: https://discuss.leetcode.com/topic/62970/simple-java-solution,
         * <p>
         * This solution does NOT modify original input.
         * Basically, it only counts the top-left one while ignoring all other parts of one battleship,
         * using the top-left one as a representative for one battle.
         * This is achieved by counting cells that don't have 'X' to the left and above them.
         */
        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] == '.') {
                        continue;//if it can pass this line, then board[i][j] must be 'X'
                    }
                    if (j > 0 && board[i][j - 1] == 'X') {
                        continue;//then we check if its left is 'X'
                    }
                    if (i > 0 && board[i - 1][j] == 'X') {
                        continue;//also check if its top is 'X'
                    }
                    count++;
                }
            }
            return count;
        }
    }

    public static class Solution2 {
        /**
         * My original solution, actually modified the input. I just undo it at the end.
         */
        public int countBattleships(char[][] board) {
            if (board == null || board.length == 0) {
                return 0;
            }
            int result = 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] == 'X') {
                        result++;
                        dfs(board, i, j, m, n);
                    }
                }
            }

            for (int i = 0; i < m; i++) {
                for (int j = 0; j < n; j++) {
                    if (board[i][j] == '#') {
                        board[i][j] = 'X';
                    }
                }
            }
            return result;
        }

        private void dfs(char[][] board, int x, int y, int m, int n) {
            if (x < 0 || x >= m || y < 0 || y >= n || board[x][y] != 'X') {
                return;
            }
            if (board[x][y] == 'X') {
                board[x][y] = '#';
            }
            dfs(board, x + 1, y, m, n);
            dfs(board, x, y + 1, m, n);
            dfs(board, x - 1, y, m, n);
            dfs(board, x, y - 1, m, n);
        }
    }

}