每日一题-26

题目描述

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：

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'

解题思路

这题简单来说，就是先找到车的位置。然后往上下左右四个地方进行遍历。

• 如果越界情况或者象，则停止此方向的遍历。
• 如果遇到p,则次数+1，且停止遍历。
  所以这题次数最大等于4，也就是往上下左右四个方向都能遇到一个卒。

代码

class Solution:
    def numRookCaptures(self, board: List[List[str]]) -> int:
        dicr = [(1,0),(-1,0),(0,-1),(0,1)]
        for i in range(8):
            for j in range(8):
                if board[i][j]=='R':
                    res = 0
                    for dx,dy in dicr:
                        x, y = i, j
                        while True:
                            x += dx
                            y += dy
                            if x>7 or x<0 or y>7 or y<0 or board[x][y]== 'B':
                                break
                            if board[x][y] == 'p':
                                res += 1
                                break
                    return res
        return 0