Availablecapturesforrook问题来源：https://leetcode-cn.com/problems/available-captures-for-rook问题是在一个8x8的棋盘上有一个白色的车（车）.也可能有空方格、白象和黑兵。它们分别由字符“R”、“.”、“B”和“p”给出。大写字母代表白色棋子，小写字母代表黑色棋子。车按照国际象棋的规则移动：它选择四个主要方向（北、东、西、南）之一，然后朝那个方向移动，直到它选择停下来，到达棋盘的边缘，或者移动到同一个广场。捕获该方格上相反颜色的棋子。此外，车不能进入与其他友好（白色）主教相同的方格。返回车一次可以捕获的兵数。例1：输入：[[".",".",".",".",".",".",".","."],[".",".","".","p",".",".",".","."],[".",".",".","R",".",".","".","p"],[".",".",".",".",".",".",".","."],[".,"”,".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".,".",".",".",".",".",".","."]]输出：3解释：在这个例子中，车可以吃掉所有的兵。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：Bishop阻止车吃掉任何兵。例3：输入：[[".",".",".",".",".",".",".","."],[".",".","".","p",".",".",".","."],[".",".",".","p",".",".",".","."],["p","p",".","R",".","p","B","."],[".",".",".",".",".",".",".","."],[".",".",".","B",".",".",".","."],[".",".",".","p",".",".",".","."],[".,".",".",".",".",".",".","."]]输出：3解释：车可以在位置b5、d6和f5吃兵。提示：棋盘.length==board[i].length==8boardi可以是'R','.','B'or'p'，只有一个格子存在boardi=='R'以上图片来源来自LeetCodeLet我稍微吐槽一下解题思路，这道题从文中理解题意，总会遇到很多麻烦（下文会提到），这里举个例子来理解：whiterook（rook,主角),白主教(bishop,friendly),blackpawn(pawn,姑且称其为敌人),必须捕获黑pawn。谜题1：选择四个基本方向（北、东、西、南）中的一个并朝该方向移动，直到它选择停止、到达棋盘边缘或移动到同一个方格以捕获该方格上的相反颜色的pawn后面的内容确实有点啰嗦（移动到同一个方格中捕获同一个方格中的相反颜色的pawn），其实就是说当车（R）被发现的时候，从车的角度出发，选择北、南、西、东（即上、下、左、右），决定可以俘获的黑兵数量。迷恋2：返回车一次可以捕获的棋子数量。从这里的文字来看，感觉车（R）选择了一个方向，一个动作俘获的兵数（p）。但是结合例子，这里的一次是指[North,South,West,East]四个方向都算一次。这里有一点需要注意，见例3，当有两个黑兵在同一个位置时，车（R）只先吃掉暴露在它前面的兵，后面的兵不会被吃掉。明白题意后，我们大致说一下解题思路：先处理边界问题，防止棋子跳出棋盘；如果遇到白棋子（B，友方），此时停止，因为你们不能同时在同一个方格；如果遇到黑棋子（p，敌方），说明你已经找到了捕获对象，返回值为+1；如果你遇到一个.（这里指的是空棋盒），表示可以继续走棋；代码实现类解决方案：defnumRookCaptures(self,board:List[List[str]])->int:defmove(board,x,y,direction):#Constraintsarewithintheboardwhilex>=0andx<8andy>=0andy<8:#遇到象时停止ifboard[x][y]=='B':break#遇到黑兵时捕获ifboard[x][y]=='p':returnTrue#继续移动x+=direction[0]y+=direction[1]returnFalsedirections=[(0,1),(1,0),(0,-1),(-1,0)]forxinrange(8):foryinrange(8):#首先定位汽车(R)ifboard[x][y]=='R':res=0#移动到四方向，方向方向：如果移动（板，x，y，direction):res+=1returnres得到结果以上就是通过分析题意，直接按照题意中陈述的规则模拟棋子的走法，解决《车的可用捕获量》问题的主要内容的问题。
