LeetCode-Python-999. 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：

LeetCode-Python-999. 车的可用捕获量

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：

LeetCode-Python-999. 车的可用捕获量

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：

简单的四方向DFS，终止条件为碰壁，吃掉P或者被B阻挡。


#dfs R不能被b挡住，目标是吃p，四个方向移动
class Solution(object):
    def numRookCaptures(self, board):
        """
        :type board: List[List[str]]
        :rtype: int
        """
        lx = 8
        ly = 8
        res = 0
        for index, item in enumerate(board):
            if "R" in item:
                Rposx, Rposy = index, item.index("R")

        posx = Rposx   
        posy = Rposy     
        while(posx > 0):
            if board[posx][posy] == "B":
                break
            elif board[posx][posy] == "p":
                res += 1
                break
            posx -= 1

        posx = Rposx   
        posy = Rposy
        while(posx < lx):
            
            if board[posx][posy] == "B":
                break
            elif board[posx][posy] == "p":
                res += 1
                break
            posx += 1

        posx = Rposx   
        posy = Rposy
        while(posy >=0):

            if board[posx][posy] == "B":
                break
            elif board[posx][posy] == "p":
                res += 1
                break
            posy -= 1

        posx = Rposx   
        posy = Rposy
        while(posy < ly):
            if board[posx][posy] == "B":
                break
            elif board[posx][posy] == "p":
                res += 1
                break
            posy += 1
        return res

LeetCode-Python-999. 车的可用捕获量

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