999-available-captures-for-rook.md

999. Available Captures for Rook - 车的可用捕获量

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'

---

题目标签：Array

题目链接：LeetCode / LeetCode中国

题解

Language	Runtime	Memory
java	0 ms	35.4 MB

class Solution {
    public int numRookCaptures(char[][] board) {
        int n = 8;
        int res = 0;
        // 找到white rook的位置
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (board[i][j] == 'R') {
                    // 从white rook的位置往4个方向进行扩展
                    for (int x = i, y = j; x >= 0; x--) {
                        if (board[x][y] == 'p') {
                            res += 1;
                            break;
                        } else if (board[x][y] == 'B') {
                            break;
                        }
                    }
                    for (int x = i, y = j; x < n; x++) {
                        if (board[x][y] == 'p') {
                            res += 1;
                            break;
                        } else if (board[x][y] == 'B') {
                            break;
                        }
                    }
                    for (int x = i, y = j; y >= 0; y--) {
                        if (board[x][y] == 'p') {
                            res += 1;
                            break;
                        } else if (board[x][y] == 'B') {
                            break;
                        }
                    }
                    for (int x = i, y = j; y < n; y++) {
                        if (board[x][y] == 'p') {
                            res += 1;
                            break;
                        } else if (board[x][y] == 'B') {
                            break;
                        }
                    }
                    break;
                }
            }
        }
        return res;
    }
}