Leetcode-419. Battleships in a Board

前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

博客链接：mcf171的博客

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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?
这个只要算当前点的上一个或者做一个位置有没有'X'，没有就加一。Your runtime beats 31.99% of java submissions.
public class Solution {
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') continue;
else if(j > 0 && board[i][j-1] == 'X') continue;
count ++;
}
}
return count;
}
}