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;
}
};