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999. 可以被一步捕获的棋子数

English Version

题目描述

在一个 8 x 8 的棋盘上，有一个白色的车（Rook），用字符 'R' 表示。棋盘上还可能存在空方块，白色的象（Bishop）以及黑色的卒（pawn），分别用字符 '.'，'B' 和 'p' 表示。不难看出，大写字符表示的是白棋，小写字符表示的是黑棋。

车按国际象棋中的规则移动。东，西，南，北四个基本方向任选其一，然后一直向选定的方向移动，直到满足下列四个条件之一：

  • 棋手选择主动停下来。
  • 棋子因到达棋盘的边缘而停下。
  • 棋子移动到某一方格来捕获位于该方格上敌方（黑色）的卒，停在该方格内。
  • 车不能进入/越过已经放有其他友方棋子（白色的象）的方格，停在友方棋子前。

你现在可以控制车移动一次，请你统计有多少敌方的卒处于你的捕获范围内（即，可以被一步捕获的棋子数）。

 

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：

  1. board.length == board[i].length == 8
  2. board[i][j] 可以是 'R'，'.'，'B' 或 'p'
  3. 只有一个格子上存在 board[i][j] == 'R'

解法

先找到 R 的位置，之后向“上、下、左、右”四个方向查找，累加结果。

Python3

class Solution:
    def numRookCaptures(self, board: List[List[str]]) -> int:

        def search(board, i, j, direction):
            while i >= 0 and i < 8 and j >= 0 and j < 8:
                if board[i][j] == 'B': return 0
                if board[i][j] == 'p': return 1
                i += direction[0]
                j += direction[1]
            return 0
        directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]
        res = 0;
        for i in range(8):
            for j in range(8):
                if board[i][j] == 'R':
                    for direction in directions:
                        res += search(board, i, j, direction)
                    return res

Java

class Solution {
    public int numRookCaptures(char[][] board) {
        int[][] directions = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
        int res = 0;
        for (int i = 0; i < 8; ++i) {
            for (int j = 0; j < 8; ++j) {
                if (board[i][j] == 'R') {
                    for (int[] direction : directions) {
                        res += search(board, i, j, direction);
                    }
                    return res;
                }
            }
        }
        return res;
    }

    private int search(char[][] board, int i, int j, int[] direction) {
        while (i >= 0 && i < 8 && j >= 0 && j < 8) {
            if (board[i][j] == 'B') return 0;
            if (board[i][j] == 'p') return 1;
            i += direction[0];
            j += direction[1];
        }
        return 0;
    }
}

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