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给定一个 8 x 8 的棋盘，只有一个 白色的车，用字符 'R' 表示。棋盘上还可能存在白色的象 'B' 以及黑色的卒 'p'。空方块用字符 '.' 表示。

车可以按水平或竖直方向（上，下，左，右）移动任意个方格直到它遇到另一个棋子或棋盘的边界。如果它能够在一次移动中移动到棋子的方格，则能够 吃掉 棋子。

注意：车不能穿过其它棋子，比如象和卒。这意味着如果有其它棋子挡住了路径，车就不能够吃掉棋子。

返回白车 攻击 范围内 兵的数量。

示例 1：

输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：3
解释：
在本例中，车能够吃掉所有的卒。

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：
象阻止了车吃掉任何卒。

示例 3：

输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","p",".",".",".","."],["p","p",".","R",".","p","B","."],[".",".",".",".",".",".",".","."],[".",".",".","B",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：3
解释： 
车可以吃掉位置 b5，d6 和 f5 的卒。

提示：

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

分析：模拟。先找到车的位置，再分别向四个方向移动直到碰到第一个棋子。若棋子是黑卒，则答案数量加1，否则不加。最后返回答案。

int numRookCaptures(char** board, int boardSize, int* boardColSize) {int ans=0;int li,lj;for(int i=0;i<boardSize;++i){int f=0;for(int j=0;j<boardSize;++j){if(board[i][j]=='R'){li=i,lj=j;break;}}}for(int j=lj-1;j>=0;j--)if(board[li][j]=='B')break;else if(board[li][j]=='p'){ans++;break;}for(int j=lj+1;j<boardSize;j++)if(board[li][j]=='B')break;else if(board[li][j]=='p'){ans++;break;}for(int i=li-1;i>=0;i--)if(board[i][lj]=='B')break;else if(board[i][lj]=='p'){ans++;break;}for(int i=li+1;i<boardSize;i++)if(board[i][lj]=='B')break;else if(board[i][lj]=='p'){ans++;break;}return ans;
}