LeetCode 419. Battleships in a Board

原题链接在这里：https://leetcode.com/problems/battleships-in-a-board/

题目：

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?

题解：

无论是1*N 还是 N*1型的battleship, 都以左上角的X来标志. 只有遇到X, 并且左边或者上边没有连着的X时才算新的battleship.

Time Complexity: O(mn), m = board.length, n = board[0].length.

Space: O(1).

AC Java:

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