Q：
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：
输入： [ [".",".",".",".",".",".",".","."], [".",".",".","p",".",".",".","."], [".",".",".","R",".",".",".","p"], [".",".",".",".",".",".",".","."], [".",".",".",".",".",".",".","."], [".",".",".","p",".",".",".","."], [".",".",".",".",".",".",".","."], [".",".",".",".",".",".",".","."]] 输出：3 解释： 在本例中，车能够捕获所有的卒。
链接：https://leetcode-cn.com/problems/available-captures-for-rook/description/
思路：笨方法
代码：
class Solution(object):
def numRookCaptures(self, board):
"""
:type board: List[List[str]]
:rtype: int
"""
row = []
col = []
for i in range(len(board)):
for j in range(len(board[0])):
if board[i][j] == 'R':
row = board[i]
col = [board[k][j] for k in range(8)]
return self.check(row) + self.check(col)
def check(self,input):
index = input.index("R")
left = input[:index]
right = input[index+1:]
count = 0
while left:
tmp = left.pop()
if tmp == "p":
count +=1
break
if tmp == 'B':
break
while right:
tmp = right.pop(0)
if tmp == "p":
count +=1
break
if tmp == 'B':
break
return count