leetcode 419甲板上的战舰

leetcode
扫描一遍
如果左边或者上边有X则不加，反之则加，注意边界判断

class Solution {
public:
    int countBattleships(vector<vector<char>>& board) {
        int num = 0;
        for(int i = 0; i < board.size(); ++i)
        {
            for(int j = 0; j < board[i].size(); ++j)
            {
                if(board[i][j]=='X')
                {
                    if(i + j == 0)
                        num++;
                    else if(i == 0 && board[i][j - 1] == '.')
                        num++;
                    else if(j == 0 && board[i - 1][j] == '.')
                        num++;
                    else if(i * j && board[i][j - 1] == '.' && board[i - 1][j] == '.')
                        num++;
                }
            }
        }
        return num;
    }
};

标签：++,419,int,num,board,&&,战舰,else,leetcode
来源： https://blog.csdn.net/qq_43409560/article/details/122032223