在一個 8 x 8 的棋盤上，有一個白色車（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則移動：它選擇四個基本方向中的一個（北，東，西和南），然后朝那個方向移動，直到它選擇停止、到達棋盤的邊緣或移動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）象進入同一個方格。

返回車能夠在一次移動中捕獲到的卒的數量。

示例 1：

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：

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：

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'

只有一個格子上存在 board[i][j] == 'R'

思路：

• 題目的意思是中間R走一下能吃p的次數，方向是上下左右四個方向

• 限制條件是走一次，以及遇到B就表明該方向不通

class Solution:    def numRookCaptures(self, board: List[List[str]]) -> int:        cnt, st, ed = 0, 0, 0        # 方向數組        direction = [(0, 1), (0, -1), (1, 0), (-1, 0)]        # 找到R        for i in range(8):            for j in range(8):                if board[i][j] == 'R':                    st = i                    ed = j        # 朝著四個方向探索        for i in range(4):            step = 0            while True:                dx, dy = direction[i]                tx = st + step * dx                ty = ed + step * dy                if tx < 0 or tx >= 8 or ty < 0 or ty >= 8 or board[tx][ty] == 'B':                    break                if board[tx][ty] == 'p':                    cnt += 1                    break                 step += 1  #每次移動的范圍，第一次移動一格，找不到的話移動兩格        return cnt