「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","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")
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