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：

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：

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：

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'

题目的理解：

题目很长，但是说的很情况，仔细看两遍应该就会明白是我们算什么值。简单描述就是：车能够吃到几个卒。

python实现

from typing import List

class Solution:
    def numRookCaptures(self, board: List[List[str]]) -> int:
        r_row = 0
        r_column = 0

        for index, row in enumerate(board):
            try:
                r_column = row.index('R')
                r_row = index
                break
            except ValueError:
                continue
        
        pawn_count = 0
        index = r_row - 1
        while index >= 0:
            value = board[index][r_column]
            if value == '.':
                index -= 1
                continue
            
            if value == 'B':
                break
            
            if value == 'p':
                pawn_count += 1
                break
                
        index = r_row + 1
        while index < 8:
            value = board[index][r_column]
            if value == '.':
                index += 1
                continue

            if value == 'B':
                break

            if value == 'p':
                pawn_count += 1
                break
        
        index = r_column - 1
        while index >= 0:
            value = board[r_row][index]
            if value == '.':
                index -= 1
                continue

            if value == 'B':
                break

            if value == 'p':
                pawn_count += 1
                break
                
        index = r_column + 1
        while index < 8:
            value = board[r_row][index]
            if value == '.':
                index += 1
                continue

            if value == 'B':
                break

            if value == 'p':
                pawn_count += 1
                break
        
        return pawn_count

看到代码又长又冗余，真的是哭了，用力想了下，没有好的办法。 当然这样的写好，很直观。

想看最优解法移步此处 有更好的编写方法。

提交

ok

// END 原来脑子是要接触尽然多的事物，它才能发挥想象力和优势