题目：

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：

image

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：

image

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：

image

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'

思路：

车只能往上下左右四个方向走，所以只需要遍历四次就可以，结束条件是i或者j等于0或者7，但是需要先遍历一遍数组找到R，所以这是一个o(n^2)
注意一个小问题，单引号' 引起的是char类型，双引号"引起的是string类型
代码如下：

class Solution {
public:
    int numRookCaptures(vector<vector<char>>& board) {
        int R_index_x;
        int R_index_y;
        int i = 0, j = 0;
        bool flag = false;
        int count = 0;
      //找到R
        for (i = 0; i < 8 && !flag; i++)
        {
            for (j = 0; j < 8; j++)
            {
                if (board[i][j] == 'R')
                {
                    R_index_x = i;
                    R_index_y = j;
                    flag = true;
                    break;
                }
            }
        }
        //往左走
        for (i = R_index_x, j = R_index_y; i >= 0; i--)
        {
            if (board[i][j] == 'p')
            {
                count++;
                break;
            }
            if (board[i][j] == 'B')
                break;
        }
        //往右走
        for (i = R_index_x, j = R_index_y; i < 8; i++)
        {
            if (board[i][j] == 'p')
            {
                count++;
                break;
            }
            if (board[i][j] == 'B')
                break;
        }
        //往上走
        for (i = R_index_x, j = R_index_y; j < 8; j++)
        {
            if (board[i][j] == 'p')
            {
                count++;
                break;
            }
            if (board[i][j] == 'B')
                break;
        }
        for (i = R_index_x, j = R_index_y; j >= 0; j--)
        {
            if (board[i][j] == 'p')
            {
                count++;
                break;
            }
            if (board[i][j] == 'B')
                break;
        }
        return count;
    }
};