LeetCode：999. 车的可用捕获量

999. 车的可用捕获量

在一个 8 x 8 的棋盘上，有一个白色车（rook）。也可能有空方块，白色的象（bishop）和黑色的卒（pawn）。它们分别以字符 “R”，“.”，“B” 和 “p” 给出。大写字符表示白棋，小写字符表示黑棋。

车按国际象棋中的规则移动：它选择四个基本方向中的一个（北，东，西和南），然后朝那个方向移动，直到它选择停止、到达棋盘的边缘或移动到同一方格来捕获该方格上颜色相反的卒。另外，车不能与其他友方（白色）象进入同一个方格。

返回车能够在一次移动中捕获到的卒的数量。

示例 1：

输入：
[[".",".",".",".",".",".",".","."],
[".",".",".","p",".",".",".","."],
[".",".",".","R",".",".",".","p"],
[".",".",".",".",".",".",".","."],
[".",".",".",".",".",".",".","."],
[".",".",".","p",".",".",".","."],
[".",".",".",".",".",".",".","."],
[".",".",".",".",".",".",".","."]]
输出：3
解释：
在本例中，车能够捕获所有的卒。

示例 2：

输入：
[[".",".",".",".",".",".",".","."],
[".","p","p","p","p","p",".","."],
[".","p","p","B","p","p",".","."],
[".","p","B","R","B","p",".","."],
[".","p","p","B","p","p",".","."],
[".","p","p","p","p","p",".","."],
[".",".",".",".",".",".",".","."],
[".",".",".",".",".",".",".","."]]
输出：0
解释：
象阻止了车捕获任何卒。

示例 3：

输入：
[[".",".",".",".",".",".",".","."],
[".",".",".","p",".",".",".","."],
[".",".",".","p",".",".",".","."],
["p","p",".","R",".","p","B","."],
[".",".",".",".",".",".",".","."],
[".",".",".","B",".",".",".","."],
[".",".",".","p",".",".",".","."],
[".",".",".",".",".",".",".","."]]
输出：3
解释： 
车可以捕获位置 b5，d6 和 f5 的卒。

提示：

1. board.length == board[i].length == 8
2. board[i][j] 可以是 'R'，'.'，'B' 或 'p'
3. 只有一个格子上存在 board[i][j] == 'R'

题解：你有你的方向数组，我有我的四个for循环，出题的用意应该是学习方向数组。

class Solution {
public:
    int numRookCaptures(vector<vector<char>>& board) {
        int res=0;
        int x=0,y=0,k=0;
        for(int i=0;i<8;i++){       
            for(int j=0;j<8;j++){
                if(board[i][j]=='R'){
                    x=i;y=j;
                    k=1;
                    break;
                }
            }
            if(k==1)break;
        }

        // 上右下左
        int dx[4]={-1,0,1,0};
        int dy[4]={0,1,0,-1};

        for(int i=0;i<4;i++){
            for(int j=1;;j++){
                int nx=x+dx[i]*j;
                int ny=y+dy[i]*j;
                if(nx<0||ny<0||nx>=8||ny>=8||board[nx][ny]=='B')break;
                if(board[nx][ny]=='p'){
                    res++;
                    break;
                }
            }
        }
        return res;
    }
};