【LeetCode】999. 车的可用捕获量

题目链接：

999. 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：

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：

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

思路：

这道题很无聊，一句话形容这道题目：老太太的裹脚布——又臭又长。

题目本身非常简单，但是题目描述让人一言难尽。

题目的意思是，棋盘中有一个“车”，问“车”向上下左右四个方向遍历能吃到多少个“卒”。条件是：

1. 不能出棋盘；
2. 遇到“象”不通；
3. 一旦吃到“卒”了，这个方向上的遍历就结束了。

一旦描述清楚，题目就变得非常简单了。也懒得想了，直接简单粗暴吧。

class Solution {
    public int numRookCaptures(char[][] board) {
        // 结果数
        int count = 0;
        // 车的坐标
        int rooki = 0, rookj = 0;
        // 找到车的位置
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[i].length; j++) {
                // 题目中已经说明只有一个“车”，所以一旦找到就可以终止寻找了。
                // 但棋盘只有8*8，跳不跳循环也没啥区别。
                if (board[i][j] == 'R') {
                    rooki = i;
                    rookj = j;
                }
            }
        }

        // 向左
        for (int j = rookj; j >= 0; j--) {
            if (board[rooki][j] == 'B') {
                break;
            }
            if (board[rooki][j] == 'p') {
                count ++;
                break;
            }
        }

        // 向右
        for (int j = rookj; j < board[rooki].length; j++) {
            if (board[rooki][j] == 'B') {
                break;
            }
            if (board[rooki][j] == 'p') {
                count ++;
                break;
            }
        }

        // 向上
        for (int i = rooki; i >= 0; i--) {
            if (board[i][rookj] == 'B') {
                break;
            }
            if (board[i][rookj] == 'p') {
                count ++;
                break;
            }
        }

        // 向下
        for (int i = rooki; i < board.length; i++) {
            if (board[i][rookj] == 'B') {
                break;
            }
            if (board[i][rookj] == 'p') {
                count ++;
                break;
            }
        }

        return count;
    }
}

原文地址：https://www.cnblogs.com/ME-WE/p/12572819.html