LeetCode--Battleships in a Board

思路:

    遍历数组,对于每个'X',判断其左边和上边是否有'X',若没有,则结果+1.

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