leetcode 999. 车的可用捕获量

题目

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。

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'
来源：leetcode
链接:https://leetcode-cn.com/problems/available-captures-for-rook/

思路

找到R，遍历就行了。用python刷题写出这么长的代码，我TM真是个小天才。

代码

class Solution:
    def numRookCaptures(self, board: List[List[str]]) -> int:
        for i in range(len(board)):
             for j in range(len(board)):
                 if board[i][j]=='R':
                    y=i
                    x=j
        count=0
        for i in range(x,-1,-1):
                if board[y][i] == 'B':
                    break
                elif board[y][i] == 'p':
                    count += 1
                    break

        for i in range(x,8):
                 if board[y][i] == 'B':
                    break
                 elif board[y][i] == 'p':
                    count += 1
                    break
        for i in range(y,-1,-1):
                if board[i][x] == 'B':
                    break
                elif board[i][x] == 'p':
                    count += 1
                    break
        for i in range(y,8):
                 if board[i][x] == 'B':
                    break
                 elif board[i][x] == 'p':
                    count += 1
                    break
        return  count