【数组】999. 可以被一步捕获的棋子数（简单）

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本文介绍了一个算法问题，目标是在一个8x8的棋盘上找到由白色车（Rook）一步可捕获的所有黑色卒的位置。文章通过示例详细解释了车的移动规则，并提供了一段C++代码实现。

【题目】
在一个 8 x 8 的棋盘上，有一个白色的车（Rook），用字符 ‘R’ 表示。棋盘上还可能存在空方块，白色的象（Bishop）以及黑色的卒（pawn），分别用字符 ‘.’，‘B’ 和 ‘p’ 表示。不难看出，大写字符表示的是白棋，小写字符表示的是黑棋。
车按国际象棋中的规则移动。东，西，南，北四个基本方向任选其一，然后一直向选定的方向移动，直到满足下列四个条件之一：
棋手选择主动停下来。
棋子因到达棋盘的边缘而停下。
棋子移动到某一方格来捕获位于该方格上敌方（黑色）的卒，停在该方格内。
车不能进入/越过已经放有其他友方棋子（白色的象）的方格，停在友方棋子前。
你现在可以控制车移动一次，请你统计有多少敌方的卒处于你的捕获范围内（即，可以被一步捕获的棋子数）。
【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】
board.length == board[i].length == 8
board[i][j] 可以是 ‘R’，’.’，‘B’ 或 ‘p’
只有一个格子上存在 board[i][j] == ‘R’
【代码】

class Solution {
public:
    int numRookCaptures(vector<vector<char>>& board) {
        int fangxiang[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
        int Rx=0,Ry=0,cnt=0;
        for(int x=0;x<board.size();x++)
            for(int y=0;y<board[0].size();y++)
                if(board[x][y]=='R'){
                    Rx=x;
                    Ry=y;
                }

        for(int c=0;c<4;c++){
            int x=Rx,y=Ry;
            for(int i=0;i<8;i++){
                x+=fangxiang[c][0];
                y+=fangxiang[c][1];
                if(x>=0&&x<8&&y>=0&&y<8){//没有越界
                    if(board[x][y]=='p'){
                        cnt++;
                        break;
                    }else if(board[x][y]=='B')
                        break;
                }
            }
        }
        return cnt;
    }
};