leetcode中如何解决车的可用捕获量问题

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'

思路：

• 题目的意思是中间R走一下能吃p的次数，方向是上下左右四个方向

• 限制条件是走一次，以及遇到B就表明该方向不通

class Solution:    def numRookCaptures(self, board: List[List[str]]) -> int:        cnt, st, ed = 0, 0, 0        # 方向数组        direction = [(0, 1), (0, -1), (1, 0), (-1, 0)]        # 找到R        for i in range(8):            for j in range(8):                if board[i][j] == 'R':                    st = i                    ed = j        # 朝着四个方向探索        for i in range(4):            step = 0            while True:                dx, dy = direction[i]                tx = st + step * dx                ty = ed + step * dy                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  #每次移动的范围，第一次移动一格，找不到的话移动两格        return cnt