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;
    }
}