Available Captures for Rook

On an 8 x 8 chessboard, there is one white rook.  There also may be empty squares, white bishops, and black pawns.  These are given as characters 'R', '.', 'B', and 'p' respectively. Uppercase characters represent white pieces, and lowercase characters represent black pieces.

The rook moves as in the rules of Chess: it chooses one of four cardinal directions (north, east, west, and south), then moves in that direction until it chooses to stop, reaches the edge of the board, or captures an opposite colored pawn by moving to the same square it occupies.  Also, rooks cannot move into the same square as other friendly bishops.

Return the number of pawns the rook can capture in one move.

Example 1:

Input: [[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
Output: 3
Explanation: 
In this example the rook is able to capture all the pawns.

Example 2:

Input: [[".",".",".",".",".",".",".","."],[".","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",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
Output: 0
Explanation: 
Bishops are blocking the rook to capture any pawn.

Example 3:

Input: [[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","p",".",".",".","."],["p","p",".","R",".","p","B","."],[".",".",".",".",".",".",".","."],[".",".",".","B",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."]]
Output: 3
Explanation: 
The rook can capture the pawns at positions b5, d6 and f5.

Note:

1. board.length == board[i].length == 8
2. board[i][j] is either 'R', '.', 'B', or 'p'
3. There is exactly one cell with board[i][j] == 'R'

题目理解：

如图，给定一张棋盘，棋盘上有一个白方的‘车’，若干个白方的‘象’和若干个黑方的‘卒’，问白方的‘车’在一步之内可以吃到多少个‘卒’

解题思路：

找到‘车’所在的位置，从这个位置往上下左右四个方向逐一搜索，搜索到‘卒’的话就记录加1，搜索到任意棋子都停止这个方向上的搜索

class Solution {
    public int numRookCaptures(char[][] board) {
        int x = -1, y = -1;
        int row = board.length, col = board[0].length;
        for(int i = 0; i < row; i++){
            for(int j = 0; j < col; j++){
                if(board[i][j] == 'R'){
                    x = i;
                    y = j;
                }
            }
        }
        int res = 0;
        for(int j = y + 1; j < col && board[x][j] != 'B'; j++){
            if(board[x][j] == 'p'){
                res++;
                break;
            }
                
        }
        for(int j = y - 1; j > -1 && board[x][j] != 'B'; j--){
            if(board[x][j] == 'p'){
                res++;
                break;
            }
        }
        for(int i = x + 1; i < row && board[i][y] != 'B'; i++){
            if(board[i][y] == 'p'){
                res++;
                break;
            }
        }
        for(int i = x - 1; i > -1 && board[i][y] != 'B'; i--){
            if(board[i][y] == 'p'){
                res++;
                break;
            }
        }
        return res;
    }
}