在一个 8 x 8 的棋盘上，有一个白色车（rook）。也可能有空方块，白色的象（bishop）和黑色的卒（pawn）。它们分别以字符 “R”，“.”，“B” 和 “p” 给出。大写字符表示白棋，小写字符表示黑棋。

车按国际象棋中的规则移动：它选择四个基本方向中的一个（北，东，西和南），然后朝那个方向移动，直到它选择停止、到达棋盘的边缘或移动到同一方格来捕获该方格上颜色相反的卒。另外，车不能与其他友方（白色）象进入同一个方格。

返回车能够在一次移动中捕获到的卒的数量。
 

示例 1：

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

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'

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/available-captures-for-rook
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

package leetCodeTest;

public class 车的可用捕获量 {
    public static void main(String[] args) {
        char[][] board = new char[][]{{'.','.','.','.','.','.','.','.'},
                {'.','.','.','.','.','.','.','.'},
                {'.','.','.','.','.','.','.','.'},
                {'.','.','.','R','.','.','.','.'},
                {'.','.','.','.','.','.','.','.'},
                {'.','.','.','.','.','.','.','.'},
                {'.','.','.','.','.','.','.','.'},
                {'.','.','.','.','.','.','.','.'}};
        int i = numRookCaptures(board);
        System.out.println("i = " + i);
    }

    /**
     * 按照题目要求设计代码即可，此类问题没有特别的思路只是考察将文字翻译成代码的能力。
     * @param board
     * @return
     */
    public static int numRookCaptures(char[][] board) {
        int i,j,flag = 0;
        for (i=0;i<board.length;i++)
            for (j=0;j<board[i].length;j++)
                if (board[i][j] == 'R'){
                    flag = findAllp(i,j,board);
                }
        return flag;

    }

    public static int findAllp(int i,int j,char[][] board){
        int temp = i;
        int flag = j;
        int count = 0;
        while (i>=0){
            if (board[i][j] == 'B')
                break;
            if (board[i][j] == 'p'){
                count++;
                break;
            }
            i--;
        }
        i = temp;
        while (i<8)
        {
            if (board[i][j] == 'B')
                break;
            if (board[i][j] == 'p'){
                count++;
                break;
            }
            i++;
        }
        i = temp;
        while (j>=0){
            if (board[i][j] == 'B')
                break;
            if (board[i][j] == 'p'){
                count++;
                break;
            }
            j--;
        }
        j = flag;
        while (j<8)
        {
            if (board[i][j] == 'B')
                break;
            if (board[i][j] == 'p'){
                count++;
                break;
            }
            j++;
        }
        return count;
    }
}