LeetCode 999. 车的可用捕获量

发布时间:2020-03-26 14:00:03发布者:Mr.Zhang阅读(105)

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:https://leetcode-cn.com/problems/available-captures-for-rook
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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;
	}
}





本文转自博客园,原文地址:https://www.cnblogs.com/izhoujie/p/12573984.html