999. 车的可用捕获量
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。
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：
象阻止了车捕获任何卒。
示例 3：
输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","p",".",".",".","."],["p","p",".","R",".","p","B","."],[".",".",".",".",".",".",".","."],[".",".",".","B",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：3
解释：
车可以捕获位置 b5，d6 和 f5 的卒。
#获取到车的位置，直接在4个方向进行遍历
class Solution:
def numRookCaptures(self, board: List[List[str]]) -> int:
if not board:
return 0
#找到车的起始位置
for i,line in enumerate(board):
for j,w in enumerate(line):
if w == "R":
x,y = i,j
count = 0
#沿着一个方向运动，直到到达边界或者象、卒
for (dx,dy) in [(-1,0),(1,0),(0,1),(0,-1)]:
new_x = x+dx
new_y = y+dy
while not (new_x < 0 or new_x>=len(board) or new_y < 0 or new_y>=len(board[0])):
if board[new_x][new_y] == "B":
break
if board[new_x][new_y] == "p":
count += 1
break
new_x += dx
new_y += dy
return count