LeetCode-Available Captures for Rook

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Description：
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:

• board.length == board[i].length == 8
• board[i][j] is either ‘R’, ‘.’, ‘B’, or ‘p’
• There is exactly one cell with board[i][j] == ‘R’

题意：在一个8x8的棋盘上，求车所能吃到的卒的数量（分为黑白两方）；

解法：
第一步：找到车所在的位置row, col
第二步：计算车所在位置的四个方向上是否可以吃到卒（中间不可以有我方的棋子）
第三步：统计吃到的卒的数量

Java

class Solution {
    public int numRookCaptures(char[][] board) {
        int res = 0;
        int row_rook = 0;
        int col_rook = 0;
        label:
        for (row_rook = 0; row_rook < board.length; row_rook++) {
            for (col_rook = 0; col_rook < board[row_rook].length; col_rook++) {
                if (board[row_rook][col_rook] == 'R') {
                    break label;
                }
            }
        }
        res += pawns(board, row_rook, col_rook, 1, 0);
        res += pawns(board, row_rook, col_rook, -1, 0);
        res += pawns(board, row_rook, col_rook, 0, 1);
        res += pawns(board, row_rook, col_rook, 0, -1);
        
        return res;
    }
    
    private int pawns(char[][] board, int row, int col, int row_step, int col_step) {
        int res = 0;
        while (row < board.length && row >= 0 && col < board[row].length && col >= 0) {
            if (board[row][col] == 'B') {
                break;
            } else if (board[row][col] == 'p') {
                res += 1;
                break;
            }
            row += row_step;
            col += col_step;
        }
        
        return res;
    }
}