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?

*/

/*

思路：先对于等于X的数组上下左右进行查询，是否存在两个方向都存在X。如果是就return 0；

然后开始进行遍历查询，查询每一个数组元素等于X，如等于X，查询他的下方跟右方，如果存在，就将他们置为“.”

*/

class Solution {
    public int countBattleships(char[][] board) {
        int count = 0;
        int rowLength = board.length-1;
        int colLength = board[0].length-1;
        for(int i=0; i<=rowLength; i++) {//判断是否有效
            for(int j=0; j<=colLength; j++) {
                if(board[i][j]=='X') {
                    if((i-1)>=0 && (j-1)>=0) {//比较上左
                        if(board[i-1][j]=='X' && board[i][j-1]=='X')
                            return 0;
                    }
                    if((i-1)>=0 && (j+1)<colLength) {//比较上右
                        if(board[i-1][j]=='X' && board[i][j+1]=='X')
                            return 0;
                    }
                    if((i+1)<rowLength && (j-1)>=0) {//比较下左
                        if(board[i+1][j]=='X' && board[i][j-1]=='X'){
                           
                            return 0;
                        }
                    }
                    if((i+1)<=rowLength && (j+1)<=colLength) {//比较下右
                        if(board[i+1][j]=='X' && board[i][j+1]=='X')
                            return 0;
                    }
                }
            }
        }
        
        for(int i=0; i<=rowLength; i++) {
            for(int j=0; j<=colLength; j++) {
                if(board[i][j]=='X') {
                    count++;
                    board[i][j] = '.';
                    if((i+1)<=rowLength) //查找下面是否存在
                        if(board[i+1][j]=='X')
                            for(int n=i+1;n<=rowLength && board[n][j]=='X'; n++)
                                board[n][j] = '.';               
                    if((j+1)<=colLength)//查找右面是否存在
                        if(board[i][j+1]=='X')
                            for(int m=j+1; m<=colLength && board[i][m]=='X'; m++)
                                board[i][m] = '.';
                }
            }
        }
        return count;
    }
}