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” 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。

样例

输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：3
解释：
在本例中，车能够捕获所有的卒。

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。

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。

注意

• 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'

算法

(暴力枚举) $O(n)$

1. 首先找到车的位置，然后分别向四个方向枚举即可。

时间复杂度

• 每个方向最多需要遍历一行或一列，故时间复杂度为 $O(n)$。

空间复杂度

• 仅需要常数的额外空间。

C++ 代码

class Solution {
public:

    bool check(int x, int y, int &ans, const vector<vector<char>>& board) {
        if (board[x][y] == 'p') {
            ans++;
            return true;
        }

        if (board[x][y] == 'B')
            return true;
        return false;
    }

    int numRookCaptures(vector<vector<char>>& board) {
        int n = board.size();
        int x = -1, y = -1;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++)
                if (board[i][j] == 'R') {
                    x = i; y = j;
                    break;
                }
            if (x != -1)
                break;
        }

        int ans = 0;
        for (int i = y + 1; i < n; i++)
            if (check(x, i, ans, board))
                break;

        for (int i = y - 1; i >= 0; i--)
            if (check(x, i, ans, board))
                break;

        for (int i = x + 1; i < n; i++)
            if (check(i, y, ans, board))
                break;

        for (int i = x - 1; i >= 0; i--)
            if (check(i, y, ans, board))
                break;
        return ans;
    }
};