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LeetCode 36. 车的可用捕获量

2020-03-26 14:09:13 阅读：233 来源： 互联网

标签：int res 捕获 36 break ++ 捕获量 board LeetCode

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：

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'

解题思路

遍历棋盘，找到车的位置

以车为中心

上下左右移动，

遇到卒，res++，再break;

遇到象，直接break；

代码如下

public class NumRookCaptures {
      public int numRookCaptures(char[][] board) {
            // 定义上下左右四个方向
            int[] dx = {-1, 1, 0, 0};
            int[] dy = {0, 0, -1, 1};
           
            for (int i = 0; i < 8; i++) {
                for (int j = 0; j < 8; j++) {
                    // 找到白车所在的位置
                    if (board[i][j] == 'R') {
                        // 分别判断白车的上、下、左、右四个方向
                        int res = 0;
                        for (int k = 0; k < 4; k++) {
                            int x = i, y = j;
                            while (true) {
                                x += dx[k];
                                y += dy[k];
                                if (x < 0 || x >= 8 || y < 0 || y >= 8 || board[x][y] == 'B') {
                                    break;
                                }
                                if (board[x][y] == 'p') {
                                    res++;
                                    break;
                                }
                            }
                        }
                        return res;
                    }
                }
            }
            return 0;
        }
}

标签：int,res,捕获,36,break,++,捕获量,board,LeetCode
来源： https://www.cnblogs.com/Transkai/p/12574027.html

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LeetCode 36. 车的可用捕获量