LeetCode 999. 车的可用捕获量

我的leetcode刷题源码[github]：https://github.com/izhoujie/algorithmcii

leetcode 999. 车的可用捕获量

题目

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：
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。

解题思路

思路1-题目较长，读懂后就简单了；

1. 遍历找到r的位置；
2. 从r的位置向四个方向上搜寻p，且p必须是首个遇到的字符，不能是b;

总结：本题的唯一难点就是题目又臭又长，读懂题比写代码要略难一点-。-

算法源码示例

package leetcode;

/**
 * @author zhoujie
 * @date 2020年3月26日 下午1:21:57 
 * @description: 999. 车的可用捕获量
 *
 */
public class leetcode_0999 {

}

class solution_0999 {
	/**
	 * @author: zhoujie
	 * @date: 2020年3月26日 下午1:38:42 
	 * @param: @param board
	 * @param: @return
	 * @return: int
	 * @description: 1-
	 *
	 */
	public int numrookcaptures(char[][] board) {
		// 四个方向的增量
		int[] x = new int[] { 1, -1, 0, 0 };
		int[] y = new int[] { 0, 0, -1, 1 };
		// 可捕获的目标数；
		int count = 0;
		for (int i = 0; i < 8; i++) {
			for (int j = 0; j < 8; j++) {
				// 找到r的位置
				if (board[i][j] == 'r') {
					// 在四个方向上探寻p
					for (int k = 0; k < 4; k++) {
						int x1 = i + x[k];
						int y1 = j + y[k];
						while (x1 > -1 && x1 < 8 && y1 > -1 && y1 < 8) {
							if (board[x1][y1] == 'b') {
								break;
							} else if (board[x1][y1] == 'p') {
								count++;
								break;
							} else {
								x1 += x[k];
								y1 += y[k];
							}
						}
					}
					return count;
				}
			}
		}
		return count;
	}
}