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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:
        x, y, n = 0, 0, 8
        for i in range(n):
            for j in range(n):
                if board[i][j] == 'R':
                    x, y = i, j
                    break
        ans = 0
        for a, b in [[0, -1], [0, 1], [-1, 0], [1, 0]]:
            i, j = x, y
            while 0 <= i + a < n and 0 <= j + b < n and board[i + a][j + b] != 'B':
                i, j = i + a, j + b
                if board[i][j] == 'p':
                    ans += 1
                    break
        return ans

Java

class Solution {
    public int numRookCaptures(char[][] board) {
        int[] pos = find(board);
        int ans = 0, n = 8;
        int[][] dirs = new int[][] {{0, -1}, {0, 1}, {-1, 0}, {1, 0}};
        for (int[] dir : dirs) {
            int x = pos[0], y = pos[1], a = dir[0], b = dir[1];
            while (
                x + a >= 0 && x + a < n && y + b >= 0 && y + b < n && board[x + a][y + b] != 'B') {
                x += a;
                y += b;
                if (board[x][y] == 'p') {
                    ++ans;
                    break;
                }
            }
        }
        return ans;
    }

    private int[] find(char[][] board) {
        int n = 8;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (board[i][j] == 'R') {
                    return new int[] {i, j};
                }
            }
        }
        return null;
    }
}

C++

class Solution {
public:
    int numRookCaptures(vector<vector<char>>& board) {
        vector<int> pos = find(board);
        int ans = 0, n = 8;
        vector<vector<int>> dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        for (auto& dir : dirs) {
            int x = pos[0], y = pos[1], a = dir[0], b = dir[1];
            while (x + a >= 0 && x + a < n && y + b >= 0 && y + b < n && board[x + a][y + b] != 'B') {
                x += a;
                y += b;
                if (board[x][y] == 'p') {
                    ++ans;
                    break;
                }
            }
        }
        return ans;
    }

    vector<int> find(vector<vector<char>>& board) {
        int n = 8;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (board[i][j] == 'R') {
                    return {i, j};
                }
            }
        }
        return {};
    }
};

Go

func numRookCaptures(board [][]byte) int {
	n := 8

	find := func() []int {
		for i := 0; i < n; i++ {
			for j := 0; j < n; j++ {
				if board[i][j] == 'R' {
					return []int{i, j}
				}
			}
		}
		return []int{}
	}

	pos := find()
	ans := 0
	dirs := [4][2]int{{0, -1}, {0, 1}, {1, 0}, {-1, 0}}
	for _, dir := range dirs {
		x, y, a, b := pos[0], pos[1], dir[0], dir[1]
		for x+a >= 0 && x+a < n && y+b >= 0 && y+b < n && board[x+a][y+b] != 'B' {
			x += a
			y += b
			if board[x][y] == 'p' {
				ans++
				break
			}
		}
	}
	return ans
}

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