LeetCode999.车的可用捕获量

题目描述:

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

在这里插入图片描述

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:

在这里插入图片描述

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:
在这里插入图片描述

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:

1.board.length == board[i].length == 8
2.board[i][j] 可以是 'R''.''B''p'
3.只有一个格子上存在 board[i][j] == 'R'


解题思路:
1.首先来分析题意,第一次看可能有点复杂,但是多看几遍之后会发现也不难
2.题目中的棋子就有四种:白车、白象、黑卒、空方块
3.再来看每个棋子的功能:白车的作用就是选定一个方向后一直走,直到吃到黑卒为止。遇到白象或边缘后便停下,而白象就是阻挡白车,位置不动,空方块可以忽略不记
4.题目要求就是求白车吃到黑卒的数量,从题目可以看出最大值就是4;因为白车吃到一个黑卒就停止了,不管黑卒后面还有没有棋
解题代码:

public class Solution {
	public int numRookCaptures(char[][] board) {
		int num = 0;
		for (int i = 0; i < board.length; i++) {
			for (int j = 0; j < board[i].length; j++) {
				if (board[i][j] == 'R') {
					for (int m = j; m < board[i].length; m++) {
						if (board[i][m] == 'B' || m == board[i].length) {
							break;
						} else if (board[i][m] == 'p') {
							num++;
							break;
						}
					}
					for (int m = j; m >= 0; m--) {
						if (board[i][m] == 'B' || m == board[i].length) {
							break;
						} else if (board[i][m] == 'p') {
							num++;
							break;

						}
					}
					for (int m = i; m < board[j].length; m++) {
						if (board[m][j] == 'B' || m == board[j].length) {
							break;
						} else if (board[m][j] == 'p') {
							num++;
							break;
						}
					}
					for (int m = i; m >= 0; m--) {
						if (board[m][j] == 'B' || m == board[j].length) {
							break;
						} else if (board[m][j] == 'p') {
							num++;
							break;

						}
					}
				}
			}
		}
		return num;
	}

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