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。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/available-captures-for-rook
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解题思路：这种题一般是用在笔试面试的第一题或者程序设计比赛的签到题，现实中的模型让你抽象的用程序解决该问题，我们使用遍历的方式，探索车周围四个方向上有没有卒并判断在车周围向四周扩散的时候有没有象，有象的话就直接退出break
public class Solution { public int numRookCaptures(char[][] board) { // 因为题目已经明确给出 board.length == board[i].length == 8，为此不做输入检查 // 定义方向数组，可以认为是四个方向向量，在棋盘问题上是常见的做法 int[][] directions = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}}; int res = 0; for (int i = 0; i < 8; i++) { for (int j = 0; j < 8; j++) { if (board[i][j] == 'R') { for (int[] direction : directions) { if (burnout(board, i, j, direction)) { res++; } } } } } return res; } /** * burnout 横冲直撞的意思（来自欧路词典） * * @param board 输入棋盘 * @param x 当前白象位置的横坐标 * @param y 当前白象位置的纵坐标 * @param direction 方向向量 * @return 消灭一个 p，就返回 true */ private boolean burnout(char[][] board, int x, int y, int[] direction) { int i = x; int j = y; while (inArea(i, j)) { // 当遇到象的时候，是友军，路被堵死，直接返回 if (board[i][j] == 'B') { break; } //当遇到卒的时候 是敌军，拿下一血（不知道一血这个词是不是这么用的） if (board[i][j] == 'p') { return true; } //横向量与横向量相加减，列向量与列向量相加减，direction[0]为横向量，direction[1]为列向量 i += direction[0]; j += direction[1]; } return false; } /** * @param i 当前位置横坐标 * @param j 当前位置纵坐标 * @return 是否在棋盘有效范围内 */ private boolean inArea(int i, int j) { return i >= 0 && i < 8 && j >= 0 && j < 8; } }
