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在一个 8 x 8 的棋盘上，有一个白色的车（Rook），用字符 ‘R’ 表示。棋盘上还可能存在空方块，白色的象（Bishop）以及黑色的卒（pawn），分别用字符 ‘.’，‘B’ 和 ‘p’ 表示。不难看出，大写字符表示的是白棋，小写字符表示的是黑棋。
车按国际象棋中的规则移动。东，西，南，北四个基本方向任选其一，然后一直向选定的方向移动，直到满足下列四个条件之一：
棋手选择主动停下来。
棋子因到达棋盘的边缘而停下。
棋子移动到某一方格来捕获位于该方格上敌方（黑色）的卒，停在该方格内。
车不能进入/越过已经放有其他友方棋子（白色的象）的方格，停在友方棋子前。
你现在可以控制车移动一次，请你统计有多少敌方的卒处于你的捕获范围内（即，可以被一步捕获的棋子数）。
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。
提示：
board.length == board[i].length == 8
board[i][j] 可以是 ‘R’，’.’，‘B’ 或 ‘p’
只有一个格子上存在 board[i][j] == ‘R’
【官解】
根据题意模拟即可：
遍历棋盘确定白色车的下标，用 (st,ed)(st,ed) 表示。
模拟车移动的规则，朝四个基本方向移动，直到碰到卒或者白色象或者碰到棋盘边缘时停止，用cnt 记录捕获到的卒的数量。
那么如何模拟车移动的规则呢？我们可以建立方向数组表示在这个方向上移动一步的增量，比如向北移动一步的时候，白色车的 x 轴坐标减 1，而 y 轴坐标不会变化，所以我们可以用 (-1, 0) 表示白色车向北移动一步的增量，其它三个方向同理。建立了方向数组，则白色车在某个方向移动 step 步的坐标增量就可以直接计算得到，比如向北移动 step 步的坐标增量即为 (-step, 0)。
class Solution:
def numRookCaptures(self, board: List[List[str]]) -> int:
cnt, st, ed = 0, 0, 0
dx, dy = [0, 1, 0, -1],[1, 0, -1, 0]
for i in range(8):
for j in range(8):
if board[i][j] == "R":
st, ed = i, j
for i in range(4):
step = 0
while True:
tx = st + step * dx[i]
ty = ed + step * dy[i]
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
return cnt
【个解】
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