999.车的可用捕获量

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。

class Solution:
    def numRookCaptures(self, board: List[List[str]]) -> int:
    	# 先存下来棋盘的规格
        l = len(board)
        # x用来记录车子能吃到的旗子的数量
        x = 0
        # m，n用来记录车子的位置
        m, n = 0, 0
        # 第一次遍历用来找车子的位置，如果找到车子即m != 0，就break
        for i in range(l):
            for j in range(l):
                if board[i][j] == 'R':
                    m = i
                    n = j
                    break
            if m != 0:
                break
        # 从车子的位置向上寻找
        for i in range(n+1):
        	# 如果是象即'B'，就停止向上的寻找
            if board[m][n-i] == 'B':
                break
            # 如果是卒即'p'，就将x+1并停止寻找
            if board[m][n-i] == 'p':
                x +=1
                break
        # 从车子的位置向下寻找
        for i in range(l-n):
            if board[m][n+i] == 'B':
                break
            if board[m][n+i] == 'p':
                x +=1
                break
        # 从车子的位置向左寻找
        for i in range(m+1):
            if board[m-i][n] == 'B':
                break
            if board[m-i][n] == 'p':
                x +=1
                break
        # 从车子的位置向右寻找
        for i in range(l-m):
            if board[m+i][n] == 'B':
                break
            if board[m+i][n] == 'p':
                x +=1
                break
        return(x)

题解：从车子的位置出发，向四个方向寻找，如果是B就停止，如果是p就加1再停止。