999. 可以被一步捕获的棋子数

为保证权益，题目请参考 999. 可以被一步捕获的棋子数(From LeetCode).

解决方案1

CPP

#include <iostream>
#include <vector>
#include <unordered_set>
#include <unordered_map>
#include <algorithm>

using namespace std;

//999

class Solution {
public:
    int numRookCaptures(vector<vector<char>> &board) {
        int x = 0;
        int y = 0;
        for (int i = 0; i < board.size(); ++i) {
            bool flag = false;
            for (int j = 0; j < board[0].size(); ++j) {
                if (board[i][j] == 'R') {
                    x = i;
                    y = j;
                    flag = true;
                    break;
                }
            }
            if (flag) {
                break;
            }
        }

        int ans = 0;

        //up
        for (int i = x - 1; i >= 0; --i) {
            if (board[i][y] == 'B') {
                break;
            } else if (board[i][y] == 'p') {
                ans++;
                break;
            }
        }

        //down
        for (int i = x + 1; i < board.size(); ++i) {
            if (board[i][y] == 'B') {
                break;
            } else if (board[i][y] == 'p') {
                ans++;
                break;
            }
        }

        //left
        for (int i = y - 1; i >= 0; --i) {
            if (board[x][i] == 'B') {
                break;
            } else if (board[x][i] == 'p') {
                ans++;
                break;
            }
        }

        //down
        for (int i = y + 1; i < board[0].size(); ++i) {
            if (board[x][i] == 'B') {
                break;
            } else if (board[x][i] == 'p') {
                ans++;
                break;
            }
        }

        return ans;
    }
};


int main() {
    return 0;
}

Python

from typing import List


class Solution:
    def numRookCaptures(self, board: List[List[str]]) -> int:
        white_car_x = -1
        white_car_y = -1

        for row_idx, row in enumerate(board):
            for col_idx, x in enumerate(row):
                if x == "R":
                    white_car_x = row_idx
                    white_car_y = col_idx

        ans = 0
        # up
        for d in range(1, 8):
            if white_car_x - d < 0:
                break
            if board[white_car_x - d][white_car_y] == "B":
                break
            elif board[white_car_x - d][white_car_y] == "p":
                ans += 1
                break

        # down
        for d in range(1, 8):
            if white_car_x + d >= 8:
                break
            if board[white_car_x + d][white_car_y] == "B":
                break
            elif board[white_car_x + d][white_car_y] == "p":
                ans += 1
                break

        # left
        for d in range(1, 8):
            if white_car_y - d < 0:
                break
            if board[white_car_x][white_car_y - d] == "B":
                break
            elif board[white_car_x][white_car_y - d] == "p":
                ans += 1
                break

        # right
        for d in range(1, 8):
            if white_car_y + d >= 8:
                break
            if board[white_car_x][white_car_y + d] == "B":
                break
            elif board[white_car_x][white_car_y + d] == "p":
                ans += 1
                break

        return ans