难度：简单

题目内容：

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。

题解：

先找到rook的位置，然后向四个方向遍历，看碰到的第一个棋子是不是pawn，是就+1。

class Solution:
    def numRookCaptures(self, board: List[List[str]]) -> int:
        r = 0
        b = False
        #找到R
        for i in range(len(board)):
            for j in range(len(board[i])):
                if board[i][j] == "R":
                    b = True
                    break
            if b :
                break
        #向下
        for i2 in range(i+1,len(board)):
            if board[i2][j]== "p":
                r += 1
                break
            elif board[i2][j] == "B":
                break
        #向上
        for i2 in range(i-1,-1,-1):
            if board[i2][j]== "p":
                r += 1
                break
            elif board[i2][j] == "B":
                break
        #向左
        for j2 in range(j-1,-1,-1):
            if board[i][j2]== "p":
                r += 1
                break
            elif board[i][j2] == "B":
                break
        
        #向右
        for j2 in range(j+1,len(board[i])):
            if board[i][j2]== "p":
                r += 1
                break
            elif board[i][j2] == "B":
                break
        
        return r

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