leetcode_999 可以被一步捕获的棋子数

要求

在一个 8 x 8 的棋盘上，有一个白色的车（Rook），用字符 'R' 表示。棋盘上还可能存在空方块，白色的象（Bishop）以及黑色的卒（pawn），分别用字符 '.'，'B' 和 'p' 表示。不难看出，大写字符表示的是白棋，小写字符表示的是黑棋。

车按国际象棋中的规则移动。东，西，南，北四个基本方向任选其一，然后一直向选定的方向移动，直到满足下列四个条件之一：

• 棋手选择主动停下来。
• 棋子因到达棋盘的边缘而停下。
• 棋子移动到某一方格来捕获位于该方格上敌方（黑色）的卒，停在该方格内。
• 车不能进入/越过已经放有其他友方棋子（白色的象）的方格，停在友方棋子前。

你现在可以控制车移动一次，请你统计有多少敌方的卒处于你的捕获范围内（即，可以被一步捕获的棋子数）。

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'

核心代码

class Solution:
    def numRookCaptures(self, board: List[List[str]]) -> int:
        x = 0
        y = 0
        cnt = 0
        for i in range(8):
            for j in range(8):
                if board[i][j] == "R":
                    x,y = i,j
        # 右
        for i in range(y + 1,8):
            if board[x][i] == "B":
                break
            if board[x][i] == "p":
                cnt += 1
                break
        # 左
        for i in range(y - 1, -1 ,-1):
            if board[x][i] == "B":
                break
            if board[x][i] == "p":
                cnt += 1
                break
        # 下
        for i in range(x+1,8):
            if board[i][y]=="B":
                break
            if board[i][y]=="p":
                cnt+=1
                break
        # 上
        for i in range(x-1,-1,-1):
            if board[i][y]=="B":
                break
            if board[i][y]=="p":
                cnt+=1
                break       
        return cnt

解题思路：此题比较简单，就是考虑我们在当前的位置上，在上下左右四个方向上能吃到的兵，我们就考虑何时停止，在合适的规则下进行终止即可。