999. 車的可用捕獲量

在一個 8 x 8 的棋盤上，有一個白色車（rook）。也可能有空方塊，白色的象（bishop）和黑色的卒（pawn）。它們分別以字符 “R”，“.”，“B” 和 “p” 給出。大寫字符表示白棋，小寫字符表示黑棋。

車按國際象棋中的規則移動：它選擇四個基本方向中的一個（北，東，西和南），然後朝那個方向移動，直到它選擇停止、到達棋盤的邊緣或移動到同一方格來捕獲該方格上顏色相反的卒。另外，車不能與其他友方（白色）象進入同一個方格。

返回車能夠在一次移動中捕獲到的卒的數量。
 

示例 1：

輸入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
輸出：3
解釋：
在本例中，車能夠捕獲所有的卒。

示例 2：

輸入：[[".",".",".",".",".",".",".","."],[".","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",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
輸出：0
解釋：
象阻止了車捕獲任何卒。

示例 3：

輸入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","p",".",".",".","."],["p","p",".","R",".","p","B","."],[".",".",".",".",".",".",".","."],[".",".",".","B",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."]]
輸出：3
解釋： 
車可以捕獲位置 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. 只有一個格子上存在 board[i][j] == 'R'

---

Code:

class Solution {
    public int numRookCaptures(char[][] board) {
        int x=0,y=0;
        p:for (int i = 0; i < 8; i++) {
            for (int j = 0; j < 8; j++) {
                if (board[i][j]=='R'){
                    x=i;
                    y=j;
                    break p;
                }
            }
        }
        int res = 0;
        //up
        for (int i = y-1; i > -1; i--) {
            if (board[x][i]=='B'){
                break;
            }else if (board[x][i]=='p'){
                res++;
                break;
            }
        }
        
        //down
        for (int i = y+1; i < 8; i++) {
            if (board[x][i]=='B'){
                break;
            }else if (board[x][i]=='p'){
                res++;
                break;
            }
        }

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

        //right
        for (int i = x+1; i < 8; i++) {
            if (board[i][y]=='B'){
                break;
            }else if (board[i][y]=='p'){
                res++;
                break;
            }
        }
        
        return res;
    }
}