数组题目：可以被一步捕获的棋子数

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题目

标题和出处

标题：可以被一步捕获的棋子数

出处：999. 可以被一步捕获的棋子数

难度

2 级

题目描述

要求

在一个 $\text{8}\times \text{8}$ 的棋盘上，有一个白色的车（$\text{Rook}$），用字符 $\text{‘R’}$ 表示。棋盘上还可能存在空方块，白色的象（$\text{Bishop}$）以及黑色的卒（$\text{pawn}$），分别用字符 $\text{‘.’}$，$\text{‘B’}$ 和 $\text{‘p’}$ 表示。不难看出，大写字符表示的是白棋，小写字符表示的是黑棋。

车按国际象棋中的规则移动。东，西，南，北四个基本方向任选其一，然后一直向选定的方向移动，直到满足下列四个条件之一：

• 棋手选择主动停下来。
• 棋子因到达棋盘的边缘而停下。
• 棋子移动到某一方格来捕获位于该方格上敌方（黑色）的卒，停在该方格内。
• 车不能进入/越过已经放有其他友方棋子（白色的象）的方格，停在友方棋子前。

你现在可以控制车移动一次，请你统计有多少敌方的卒处于你的捕获范围内（即，可以被一步捕获的棋子数）。

示例

示例 1：

输入：$\text{[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]}$
输出：$\text{3}$
解释：
在本例中，车能够捕获所有的卒。

示例 2：

输入：$\text{[[".",".",".",".",".",".",".","."],[".","p","p","p","p","p",".","."],[".","p","p","B","p","p",".","."],[".","p","B","R","B","p",".","."],[".","p","p","B","p","p",".","."],[".","p","p","p","p","p",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]}$
输出：$\text{0}$
解释：
象阻止了车捕获任何卒。

示例 3：

输入：$\text{[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","p",".",".",".","."],["p","p",".","R",".","p","B","."],[".",".",".",".",".",".",".","."],[".",".",".","B",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."]]}$
输出：$\text{3}$
解释：
车可以捕获位置 $\text{b5}$，$\text{d6}$ 和 $\text{f5}$ 的卒。

数据范围

• $\text{board.length}=\text{board[i].length}=\text{8}$
• $\text{board[i][j]}$ 可以是 $\text{‘R’}$，$\text{‘.’}$，$\text{‘B’}$ 或 $\text{‘p’}$
• 只有一个格子上存在 $\text{board[i][j]}=\text{‘R’}$

解法

思路和算法

首先需要找到白色的车在棋盘上的位置，即所在的行和列。

然后需要判断白色的车分别往四个方向移动时，是否可以捕获黑色的卒。由于只能移动一次，因此每个方向最多可以捕获一个黑色的卒，可以捕获的黑色的卒的总数不会超过 $4$ 个。对于每个方向，如果遇到了黑色的卒，则将其捕获，然后停止移动，如果遇到了白色的象或棋盘边缘，则停止移动。

代码

class Solution {public int numRookCaptures(char[][] board) {int rookRow = -1, rookColumn = -1;int rows = board.length, columns = board[0].length;int totalSquares = rows * columns;for (int i = 0; i < totalSquares; i++) {int row = i / columns, column = i % columns;if (board[row][column] == 'R') {rookRow = row;rookColumn = column;break;}}int count = 0;int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};for (int[] direction : directions) {int tempRow = rookRow + direction[0], tempColumn = rookColumn + direction[1];while (tempRow >= 0 && tempRow < rows && tempColumn >= 0 && tempColumn < columns) {if (board[tempRow][tempColumn] != '.') {if (board[tempRow][tempColumn] == 'p') {count++;}break;}tempRow += direction[0];tempColumn += direction[1];}}return count;}
}

复杂度分析

• 时间复杂度：$O(n^2)$，其中 $n$ 是棋盘的边长，这道题中 $n=8$。
  寻找白色的车需要遍历棋盘，时间复杂度是 $O(n^2)$。
  在四个方向判断有多少敌方的卒可以被一步捕获，最多需要遍历除了车所在的格子之外的 $2n-2$ 个格子，时间复杂度是 $O(n)$。
  总时间复杂度是 $O(n^2)$。

• 空间复杂度：$O(1)$。

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