【Leetcode_总结】999. 车的可用捕获量 - python

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本文探讨了在8x8的棋盘上，一个白色车如何捕获黑色卒的问题。通过解析车的移动规则和捕获机制，文章提供了一种算法解决方案，并详细解释了其工作原理。

Q：

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。

链接：https://leetcode-cn.com/problems/available-captures-for-rook/description/

思路：笨方法

代码：

class Solution(object):
    def numRookCaptures(self, board):
        """
        :type board: List[List[str]]
        :rtype: int
        """
        row = []
        col = []
        for i in range(len(board)):
            for j in range(len(board[0])):
                if board[i][j] == 'R':
                    row = board[i]
                    col = [board[k][j] for k in range(8)]
                    return self.check(row) + self.check(col)
    
    def check(self,input):
        index = input.index("R")
        left = input[:index]
        right = input[index+1:]
        count = 0
        while left:
            tmp = left.pop()
            if tmp == "p":
                count +=1
                break
            if tmp == 'B':
                break
                
        while right:
            tmp = right.pop(0)
            if tmp == "p":
                count +=1
                break
            if tmp == 'B':
                break
        return count