題目：

在一個(gè) 8 x 8 的棋盤上，有一個(gè)白色車（rook）。也可能有空方塊，白色的象（bishop）和黑色的卒（pawn）。它們分別以字符 “R”，“.”，“B” 和 “p” 給出。大寫字符表示白棋，小寫字符表示黑棋。

車按國際象棋中的規(guī)則移動(dòng)：它選擇四個(gè)基本方向中的一個(gè)（北，東，西和南），然后朝那個(gè)方向移動(dòng)，直到它選擇停止、到達(dá)棋盤的邊緣或移動(dòng)到同一方格來捕獲該方格上顏色相反的卒。另外，車不能與其他友方（白色）象進(jìn)入同一個(gè)方格。

返回車能夠在一次移動(dòng)中捕獲到的卒的數(shù)量。

示例 1：

image

輸入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
輸出：3
解釋：
在本例中，車能夠捕獲所有的卒。
示例 2：

image

輸入：[[".",".",".",".",".",".",".","."],[".","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",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
輸出：0
解釋：
象阻止了車捕獲任何卒。
示例 3：

image

輸入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","p",".",".",".","."],["p","p",".","R",".","p","B","."],[".",".",".",".",".",".",".","."],[".",".",".","B",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."]]
輸出：3
解釋：
車可以捕獲位置 b5，d6 和 f5 的卒。

提示：

board.length == board[i].length == 8
board[i][j] 可以是 'R'，'.'，'B' 或 'p'
只有一個(gè)格子上存在 board[i][j] == 'R'

思路：

車只能往上下左右四個(gè)方向走，所以只需要遍歷四次就可以，結(jié)束條件是i或者j等于0或者7，但是需要先遍歷一遍數(shù)組找到R，所以這是一個(gè)o(n^2)
注意一個(gè)小問題，單引號(hào)' 引起的是char類型，雙引號(hào)"引起的是string類型
代碼如下：

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;
    }
};