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LeetCode: 999. 车的可用捕获量

1、题目描述

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”“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",".","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'

2、解题思路

  • 找到车(rook)的位置,然后上下左右扫描
class Solution:
    def numRookCaptures(self, board: List[List[str]]) -> int:
        R, C = len(board), len(board[0])

        rook_row = 0
        rook_col = 0

        for r, row in enumerate(board):
            for c, pos in enumerate(row):
                if pos == "R":
                    rook_row = r
                    rook_col = c

        count = 0

        # right
        for i in range(rook_row + 1, R):
            if board[i][rook_col] == "B":
                break
            elif board[i][rook_col] == "p":
                count += 1
                break
        # left
        for i in range(rook_row, -1, -1):
            if board[i][rook_col] == "B":
                break
            elif board[i][rook_col] == "p":
                count += 1
                break
        # up
        for i in range(rook_col + 1, C):
            if board[rook_row][i] == "B":
                break
            elif board[rook_row][i] == "p":
                count += 1
                break
        # down
        for i in range(rook_col - 1, -1, -1):
            if board[rook_row][i] == "B":
                break
            elif board[rook_row][i] == "p":
                count += 1
                break

        return count