「LeetCode每日一题」——999. 车的可用捕获量

`999. 车的可用捕获量`

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

`题目`

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",".","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`

开局一堆字加图，表达的意思其实很简单。车只能四个方向直走，只能吃掉敌人，友军会挡路。

我们需要的是先找到自己的车(R)在哪里，然后大杀四方，看看遇到了多少次敌人(p)。

开始循环找到R的位置，然后把X，Y轴上的所有障碍"."都清空，这样我们的车R就直接和敌人p交手了。敌人p可能在R的两侧，所有需要分别对"pR"和"Rp"进行计数，统计出与敌人的交手次数，当然车是百战百胜的，这样就得出了杀敌数。

`方案代码`

解决方案：

class Solution:
    def numRookCaptures(self, board: List[List[str]]) -> int:
        for i in range(len(board)):
            if 'R' in board[i]:
                raw = i
                break
        colum = board[raw].index('R')

        r = ''.join(board[raw]).replace('.','')
        c = ''.join(i[colum] for i in board).replace('.','')

        return r.count("Rp") + r.count("pR") + c.count("Rp") + c.count("pR")

「LeetCode每日一题」——999. 车的可用捕获量

999. 车的可用捕获量

题目

思路

方案代码

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`999. 车的可用捕获量`

`题目`

`思路`

`方案代码`