999. 可以被一步捕获的棋子数（C++）

• 1 题目描述
• 2 示例描述
• • 2.1 示例 2
  • 2.2 示例 2
  • 2.3 示例 3
• 3 解题提示
• 4 解题思路
• 5 源码详解（C++）

1 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）。

2 示例描述

2.1 示例 2

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

2.2 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。

2.3 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。

3 解题提示

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’

4 解题思路

采用两个数组将步骤叠装，后控制循环走棋。

5 源码详解（C++）

class Solution {
public:
    int numRookCaptures(vector<vector<char>>& board) {
        int n = 0 , x = 0 , y = 0 ;
        int dx[4] = { 0, 1, 0, -1 } ; //控制x方向
        int dy[4] = { 1, 0, -1, 0 } ; //控制y方向
        for ( int i = 0 ; i < 8 ; i ++ )
        {
            for ( int j = 0 ; j < 8 ; j ++ )
            {
                if ( board[i][j] == 'R' )
                {
                    x = i ;
                    y = j ;
                    break ;
                }
            }
        }

        for ( int i = 0 ; i < 4 ; i ++ ) 
        {
            /*
            i = 0 : 向右移动
            i = 1 : 向下移动
            i = 2 : 向左移动
            i = 3 : 向上移动
            */
            for ( int step = 0 ; ; step ++ ) 
            {
                int now_x = x + step * dx[i] ;
                int now_y = y + step * dy[i] ;
                if ( now_x < 0 || now_x >= 8 || now_y < 0 || now_y >= 8 || board[now_x][now_y] == 'B') //出现任意一种情况就退出循环
                {
                    break;
                }
                if ( board[now_x][now_y] == 'p' ) 
                {
                    n ++ ;
                    break ;
                }
            }
        }
        return n ;
    }
};