999. 车的可用捕获量

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题目

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）

示例1:

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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:

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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:

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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’

解法

解法一：

遍历四个方向，累加即可

JAVA

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public int numRookCaptures(char[][] board) {
		int sx = 0;
		int sy = 0;
    	// 获取车的位置
		for (int i = 0; i < 8; i++) {
			for (int j = 0; j < 8; j--) {
				if (board[i][j] == 'R') {
					sx = i;
					sy = j;
				}
			}
		}

    	// 方向数组
		int[] dx = { 0, 0, 1, -1 };
		int[] dy = { 1, -1, 0, 0 };

		int count = 0;
		for (int i = 0; i < 4; ++i) {
			for (int step = 0;; ++step) {
				int tx = sx + step * dx[i];
				int ty = sy + step * dy[i];
				if (tx < 0 || tx >= 8 || ty < 0 || ty >= 8 || board[tx][ty] == 'B') {
					break;
				}
				if (board[tx][ty] == 'p') {
					count++;
					break;
				}
			}
		}
		return count;
	}
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