题目描述：
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：
输入：[[".",".",".",".",".",".",".","."],
[".",".",".","p",".",".",".","."],
[".",".",".","R",".",".",".","p"],
[".",".",".",".",".",".",".","."],
[".",".",".",".",".",".",".","."],
[".",".",".","p",".",".",".","."],
[".",".",".",".",".",".",".","."],
[".",".",".",".",".",".",".","."]]
输出：3
解释：
在本例中，车能够捕获所有的卒。
代码实现：
# 方法一：
def numRookCaptures1(self, board: List[List[str]]) -> int:
"找到 R 所在行和列的所有元素进行切片和遍历，"
row = None
col = None
for x in range(len(board)):
if "R" in board[x]:
row = x
break
col = board[row].index("R")
count = 0
for i in board[row][col + 1 :]:
if i == ".":
pass
elif i == "B":
break
elif i == "p":
count += 1
break
for i in board[row][col - 1 :: -1]:
if i == ".":
pass
elif i == "B":
break
elif i == "p":
count += 1
break
s = ''.join(i[col] for i in board)
pos = s.index("R")
for x in s[pos + 1:]:
if x == ".":
pass
elif x == "B":
break
elif x == "p":
count += 1
break
for x in s[pos - 1::-1]:
if x == ".":
pass
elif x == "B":
break
elif x == "p":
count += 1
break
print(count)
# 方法二：
def numRookCaptures2(self, board: List[List[str]]) -> int:
'''
首先找到车所在的行和列
然后根据所在的行和列匹配车所能捕获的卒数
'''
output = 0
col = None
row = None
for i in range(len(board)):
if 'R' in board[i]:
row = i
break
col = board[row].index('R')
s = ''.join(board[row])
s = s.replace('.', '')
if 'pR' in s:
output += 1
if 'Rp' in s:
output += 1
s = ''.join([i[col] for i in board])
s = s.replace('.', '')
if 'pR' in s:
output += 1
if 'Rp' in s:
output += 1
return output
