LeetCode 419. Battleships in a Board

Given an 2D board, count how many different 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.

阅读题目,可以得知一下几个信息:

1.所谓的BattleShip其实就是x.

2.每个BattleShip之间会有.分割..

3.每个BattleShip应该是1*n个X或者是n*1的x组成的...

所以算法如下:

public class Solution {
public int countBattleships(char[][] board) {
int sum = 0;
int a = board.length;
int b = board[0].length;
if(a == 0 || b == 0)
return 0;
for(int i=0;i<a;i++)
{
for(int j=0;j<b;j++)
{
if(board[i][j] == '.')
continue;
if(i>0 && board[i-1][j] == 'X')
continue;
if(j>0 && board[i][j-1] == 'X')
continue;
sum++;
}
}
return sum;
}
}只需要排除非法选项就可以了.
1.当字符为.的时候

2.当i>0,board[i][j] = 'X' 并且board[i-1][j] ='X'时,也就是竖方向上有相连的X,这种情况下看做是同一个BattleShip

3.当j>0,board[i][j] = 'X' 并且board[i][j-1] = 'X'时,也就是横方向上有相连的X,也看做同一个...

除此之外的所有情况都是合法的.return sum就可以了.